DIT 111 Probability and queing theory.pdf

M.Sc.Information Technology (DISTANCE MODE) DIT 111 Probability and Queuing Theory I SEMESTER COURSE MATERIAL Centre for Distance Education Anna University Chennai Chennai – 600 025 Author Mrs. Maya Joby Lecturer Department of Mathematics DMI College of Engineering Chennai – 602 103 Reviewer Dr.M.Srinivasan Professor & Controller of Examinations SSN College of Engineering, Old Mahabalipuram Road, Kalavakkam Chennai. - 603110 Editorial Board Dr. C. Chellappan Professor Department of Computer Science and Engineering Anna University Chennai Chennai – 600 025 Dr. T.V. Geetha Professor Department of Computer Science and Engineering Anna University Chennai Chennai – 600 025 Dr. H. Peeru Mohamed Professor Department of Management Studies Anna University Chennai Chennai – 600 025 Copyrights Reserved (For Private Circulation only) ACKNOWLEDGEMENT I, Ms. Maya Joby, express my sincere thanks and deep sense of gratitude to the following persons who have constantly encouraged and supported me in this wonderful Endeavour. Dr. Chellappan – Professor. Department of Computer Science and Engineering. Deputy Director, Centre for Distance Education Anna University For guiding and encouraging me during this course material preparation. Dr. M.Gopal – Principal DMI College of Engineering, Chennai for the institutional support rendered and encouraging me during the preparation of this course material. All staff and authorities of Centre for Distance Education Anna University for providing me this opportunity to prepare this course material. While preparing this material I have benefited immensely by referring to many books. I express my gratitude to all such authors and publishers. Reference Books: 1). T. Veerarajan, “Probability, Statistics and Random Process” Tata McGraw Hill. 2) P. Kandasamy, K. Thilagavathi and K Gunavathi “ Probability Random variables and Random processors”. S Chand. 3) S.C. Gupta and V K Kapoor, “Fundamentals of Mathematical Statistics” Sultan Chand & Sons. I wish to acknowledge my sincere thanks to my colleagues in DMI college of Engineering for guiding and helping me whenever I was stuck in trying to better explain some topics. I also wish to thank my family members for their moral support and encouragement while preparing this material. Mrs. Maya Joby Lecturer Department of Mathematics DMI College of Engineering Chennai – 602 103 DIT 111 PROBABILITY AND QUEUING THEORY UNIT I Probability and Random Variables: Probability concepts – Random variables – Moment generating function – Binomial, Poisson, Geometric, Uniform exponential, Normal distributions – Functions of Random variables. UNIT II Two-Dimensional Random Variables: Marginal and conditional distributions – Covariance – correlation and Regression – Transformation of Random Variables. UNIT III Tests of Hypothesis: Sampling distributions – Tests based on Normal, t and F distributions for means, variance and proportions, chi-square test for variance, independence and goodness of fit. UNIT IV Random Process: Classification – stationary process – Markov process – Poisson process – Markov chains. UNIT V Queueing Theory: Single and multi-server Markovian Queues – Stationary for queue size distributions – Little’s formula – Average measures. TEXT BOOK 1. T.Veerarajan, “Probability, Statistics and Random Process, Tata McGraw Hill, 2002. REFERENCES 1. Taha, H.A. “Operations Research : An Introduction”, Prentice Hall, New Delhi, 2002. 2. P. Kandasamy, K. Thilagavathi and K. Gunavathi, “Probability, Random Variables and Random Processors”, S. Chand, 2003. DIT 111 PROBABILITY AND QUEUEING THEORY Page Nos. UNIT 1 - PROBABILITY AND RANDOM VARIABLES 1.1 INTRODUCTION 1 1.2 LEARNING OBJECTIVES 2 1.3 PROBABILITY CONCEPTS 2 1.4 RANDOM VARIABLES 21 1.5 EXPECTATION AND MOMENTS 34 1.6 MOMENT GENERATING FUNCTION 35 1.7 DISCRETE DISTRIBUTIONS 42 1.8 CONTINUOUS DISTRIBUTION 68 1.9 FUNCTIONS OF RANDOM VARIABLES 97 UNIT 2 - TWO DIMENSIONAL RANDOM VARIABLES 2.1 INTRODUCTION 101 2.2 LEARNING OBJECTIVES 101 2.3 TWO DIMENSIONAL RANDOM VARIABLES 102 2.4 A) MARGINAL PROBABILITY DISTRIBUTION 103 2.4 B) CONDITIONAL PROBABILITY DISTRIBUTION 104 2.5 EXPECTATION OF A FUNCTION 115 2.6 COVARIANCE 117 2.7 CORRELATION 121 2.8 REGRESSION 140 2.9 TRANSFORMATION OF RANDOM VARIABLES 153 UNIT 3 - TESTING OF HYPOTHESES 3.1 INTRODUCTION 161 3.2 LEARNING OBJECTIVES 161 3.3 TEST BASED ON NORMAL DISTRIBUTION 168 DIT 111 PROBABILITY AND QUEUEING THEORY Page Nos. 3.4 STUDENT’S T- DISTRIBUTION 190 3.5 VARIANCE RATIO TEST OR F-TEST 209 3.6 CHI SQUARE TEST 220 UNIT 4 - RANDOM PROCESSES 4. 1 INTRODUCTION 245 4.2 LEARNING OBJECTIVES 245 4.3 RANDOM PROCESS 246 4.4 CLASSIFICATION 246 4.5 STATIONARITY 248 4.6 MARKOV PROCESS AND MARKOV CHAIN 255 4.7 POISSON PROCESS 270 UNIT 5 - QUEUEING THEORY 5.1 INTRODUCTION 283 5.2 LEARNING OBJECTIVES 283 5.3 BASIC CHARACTERISTIC OF QUEUEING PHENOMENA 283 5.4 OPERATING CHARACTERISTICS OF QUEUEING SYSTEM 285 5.5 KENDALL’S NOTATION FOR REPRESENTING QUEUEING MODELS 286 5.6 DIFFERENCE EQUATION RELATED TO POISSON QUEUE SYSTEM 287 5.7 CHARACTERISTICS OF INFINITE CAPACITY, SINGLE SERVER POISSON QUEUE MODEL I 290 5.8 CHARACTERISTICS OF INFINITE CAPACITY, MULTIPLE SERVER POISSON QUEUE MODEL II 295 5.9 CHARACTERISTICS OF FINITE CAPACITY, SINGLE SERVER POISSON QUEUE MODEL III 300 5.10 CHARACTERISTICS OF FINITE CAPACITY, SINGLE SERVER POISSON QUEUE MODEL IV 303 DIT 111 PROBABILITY AND QUEUEING THEORY 1 NOTES Anna University Chennai UNIT 1 PROBABILITY AND RANDOM VARIABLES - Introduction - Probability Concepts - Random Variables - Expectation and Moments - Moment Generating Functions - Binomial Distribution - Poisson Distribution - Geometric Distribution - Uniform Distribution - Exponential Distribution - Normal Distribution - Functions of Random Variable 1.1 INTRODUCTION Probability theory is an important part of contemporary mathematics. It plays a key role in the insurance industry, in the modeling of financial markets, and in statistics generally — including all those fields of endeavor to which statistics is applied (e.g. health, physical sciences, engineering, economics). The 20th century has been an important period for the subject, because we have witnessed the development of a solid mathematical basis for the study of probability, especially from the Russian school of probability under the leadership of A N Kolmogorov. We have also seen many new applications of probability — from applications of stochastic calculus in the financial industry to Internet gambling. At the beginning of the 21st century, the subject offers plenty of scope for theoretical developments, modern applications and computational problems. There is something for everyone in probability! DIT 111 PROBABILITY AND QUEUEING THEORY 2 NOTES Anna University Chennai 1.2 LEARNING OBJECTIVES The students will acquire - Knowledge to define experiment, outcome, event, probability and equally likely. - State the formula for finding the probability of an event. - Knowledge to evaluate outcomes and probabilities for several simple experiments. - Knowledge to Recognize the difference between outcomes that are, and are not, equally likely to occur. - Knowledge to apply basic probability principles to solve problems. - Knowledge to analyze each problem to identify the given information. - Knowledge to identify the concepts and procedures needed to solve each problem. - Knowledge to apply probability concepts to solve complex problems. - Knowledge to identify connections between probability and the real world. - Develop strong problem-solving skills. - Familiarity with some of the distributions commonly used to represent real-life situations. First of all let us go through some basic concepts in probability. Try to be thorough in the basics so you may feel comfortable with the subject. 1.3 PROBABILITY CONCEPTS 1.3.1 Random Experiment If an experiment is repeated under the same conditions , any number of times, it does not give unique results but may result in any one of several outcomes. Thus an action which can produce any result or outcome is called a RANDOM EXPERIMENT. When a random experiment is performed each time it is called a TRIAL and the outcomes are known as EVENTS or CASES. An experiment whose outcome or result can be predicted is called a Deterministic Experiment. Sample Space : The sample space is an exhaustive list of all the possible outcomes of an experiment. Each possible result of such a study is represented by one and only one point in the sample space, which is usually denoted by S. Examples Experiment of Rolling a die once: DIT 111 PROBABILITY AND QUEUEING THEORY 3 NOTES Anna University Chennai Sample space S = {1,2,3,4,5,6} Experiment Tossing a coin: Sample space S = {Heads,Tails} Experiment Measuring the height (cms) of a girl on her first day at school: Sample space S = the set of all possible real numbers An event whose occurrence is inevitable when an experiment is performed is called as Certain event or Sure event. An event which can never occur when an experiment is performed is called an Impossible event. Events may be ‘simple’ or ‘compound’. An event is called simple if it corresponds to a single possible outcome of the experiment otherwise it is called compound event or composite event. For example: In drawing cards from a pack of 52 cards, the chance of getting spade 5 is a simple event and the chance of getting a king is compound event. Occurrence of getting two 8 diamond cards is an impossible event. Favorable Events: The number of cases favorable to an event in a trial is the number of outcomes which entail the happening of the event. Equally likely events: The outcomes are said to be equally likely if none of them is expected to occur in preference to other. Thus two or more events are said to be equally likely if each one of them has an equal chance of happening. For example : when a coin is thrown, the head is as likely to turn up as tail. Hence H and T are equally likely events. Mutually exclusive events or Incompatible events: If event A happens, then event B cannot, or vice-versa. The two events “it rained on Tuesday” and “it did not rain on Tuesday” are mutually exclusive events. Both cannot happen in a single trial or we can say that the occurrence of any one of them excludes the occurrence of other. Formally, two events A and B are mutually exclusive if and only if A1B = φ Exhaustive events: Outcomes are said to be exhaustive when they include all possible outcomes. For example, while rolling a die, the possible outcomes are 1,2,3,4,5 and 6 and hence the exhaustive number of cases is 6. Independent Events: Two events are independent if the occurrence of one of the events gives us no information about whether or not the other event will occur; that is, the events have no influence on each other. For example, if a coin is thrown twice, the result of the second throw is no way affected by the result of the first throw. Thus the events are independent events. DIT 111 PROBABILITY AND QUEUEING THEORY 4 NOTES Anna University Chennai Dependent events: Two events are said to be dependent if the occurrence or nonoccurrence of an event in any trial affects the occurrence of the other event in other trials. Complementary events: A is called complementary event of B if A and B are mutually exclusive and exhaustive. When a die is thrown, occurrence of an even number and odd number are complementary events. 1.3.2 Probability of an event : Classical definition: The outcomes of a random experiment are termed as events. The probability for the occurrence of an event A is defined as the ratio between the number of favorable outcomes for the occurrences of the event and the total number of possible outcomes. P(A) = No: of favorable cases Total no: of cases Also the probability of non-happening of event A is P(Ā) = 1 – m/n i.e, P(Ā) = 1 – P(A). Note: If P(A) = 1 the event A is called a certain event and if P(A) = 0, the event is called an impossible event. Also P(A) + P(Ā) = 1 Examples 1. The probability of drawing a spade from a pack of 52 well-shuffled playing cards is 13/52 = 1/4 = 0.25 since event E = ‘a spade is drawn’; the number of outcomes corresponding to E = 13 (spades); the total number of outcomes = 52 (cards). 1.3.2.1. Statistical definition: If a trial results in ‘n’ cases and m of them are favorable to the happening of the event then P(A) = lim (m/n) n  · 1.3.2.2 Axiomatic approach to probability: Definition of probability : Let S be a sample space associated with an experiment. To each event A, there is a real number P(A) associated , called the probability of A satisfying the following axioms: DIT 111 PROBABILITY AND QUEUEING THEORY 5 NOTES Anna University Chennai i) P(A) ¡ 0 ii) P(S) = 1 iii) If A 1 ,A 2 ,A 3 ,------A n are mutually exclusive events, then P(A 1 U A 2 U A 3 ------ U An) = P(A 1 ) + P(A 2 ) + --------P(A n ) Theorem : The probability of an impossible event is zero, i.e, if Φ is the subset (event) containing no sample point , P (Φ ) = 0 Proof – The certain event S and the impossible event Φ are mutually exclusive Hence P (S U Φ ) = P (S) + P (Φ ) [ Axiom (iii)] But S U Φ = S Thus P (S ) = P(S) + P(Φ ) Thus P(Φ ) = 0 Theorem : If Ā is the complimentary event of A, P ( Ā) = 1 – P (A ) ,1 Proof : A and Ā are mutually exclusive events, such that A U Ā = S Thus P (A U Ā) = P (S) = 1 [ Axiom (ii)] i.e. P(A) + P(Ā) = 1 [ Axiom (iii)] Thus P(Ā ) = 1 – P (A) Since P(A) ¡ 0, It follows that P( Ā ) , 1. Theorem : If A and B are any 2 events, P (A U B ) = P(A) + P(B) – P(A1B ) = P(A) + P(B) Proof : A is the union of the mutually exclusive events AB and AB and B is the union of the mutually exclusive events ĀB and AB Thus P(A) = P(AB) + P(AB ) [ Axiom (iii) ] and P(B) = P( Ā B) + P(AB) [ Axiom (iii) ] AB AB ĀB A B DIT 111 PROBABILITY AND QUEUEING THEORY 6 NOTES Anna University Chennai Thus P(A) + P(B) = [P(AB) + P(AB) + P(ĀB) + P(AB)] = P(A U B ) + P (A1B) This result follows. Clearly, P(A) +P(B) – P ( A1B) = P(A) + P(B) Theorem : If B C A, P(B) , P(A) Proof: B and AB are mutually exclusive events such that B U AB = A Thus P(B U A B ) = P(A) i.e. P(B) + P AB ) = P(A) { Axiom (iii) ] Thus P(B) = P(A) Note: The above theorem can be extended as P( A U B U C) = P (at least one A, B anc C occurs) = P(A) +P(B) + P(C) - P (A1B) - P (B1C) - P (C1A) + P (A1B1C) 1.3.3 Laws of Probability 1.3.3.1 Addition Rule : The addition rule is a result used to determine the probability that event A or event B occurs or both occur. The result is often written as follows, using set notation: P(A U B) = P(A) +P(B) – P(A1B) where: P(A) = probability that event A occurs S B AB A DIT 111 PROBABILITY AND QUEUEING THEORY 7 NOTES Anna University Chennai P(B) = probability that event B occurs P(A U B) = probability that event A or event B occurs P(A1B ) = probability that event A and event B both occur For mutually exclusive events, that is events which cannot occur together: P(A1B) = 0 The addition rule therefore reduces to P(AUB) = P(A) + P(B) For independent events, that is events which have no influence on each other: P(A1B) = P(A)P(B) The addition rule therefore reduces to P(A U B) = P(A) +P(B) – P(A)P(B) Example Suppose we wish to find the probability of drawing either a king or a spade in a single draw from a pack of 52 playing cards. We define the events A = ‘draw a king’ and B = ‘draw a spade’ Since there are 4 kings in the pack and 13 spades, but 1 card is both a king and a spade, we have: P(A U B) = P(A) +P(B) – P(A1B) = 4/52 + 13/52 - 1/52 = 16/52 So, the probability of drawing either a king or a spade is 16/52 (= 4/13). Theorem: Additive law of probability: If A and B are any two events (subsets of sample space S) are not disjoint, then P(A U B) = P(A) +P(B) – P(A1B) Proof: We have A U B = A U (Ā1B) Since A and (Ā1B) are disjoint, P(A U B) = P(A) + P(Ā1B) = P(A) +[P(Ā1B) + P(A1B)] – P(A1B) = P(A) +P[(Ā1B) U (A1B)] – P(A1B) = P(A) + P(B) – P(A1B)(since Ā1B and A1B are disjoint ) DIT 111 PROBABILITY AND QUEUEING THEORY 8 NOTES Anna University Chennai 1.3.3.2 Multiplication Rule The multiplication rule is a result used to determine the probability that two events, A and B, both occur. The Conditional Probability of an event B, assuming that the event A has happened, is denoted by P(A/B) and defined as P(B/A) = P( A1B) / P(A) The multiplication rule follows from the definition of conditional probability. The result is often written as follows, using set notation: P(A1B) = P(A/B)P(B) OR P(A1B) = P(B/A)P(A) where: P(A) = probability that event A occurs P(B) = probability that event B occurs P(A1B) = probability that event A and event B occur P(A | B) = the conditional probability that event A occurs given that event B has occurred already P(B | A) = the conditional probability that event B occurs given that event A has occurred already For independent events, that is events which have no influence on one another, the rule simplifies to: P(A1B) = P(A)P(B) OR P(A and B ) = P(A).P(B) That is, the probability of the joint events A and B is equal to the product of the individual probabilities for the two events. Theorem: Multiplication Law of Probability or Theorem of compound probabilities. For two events A and B, P(A1B) = P(A/B)P(B) , P(A)>0 = P(B/A)P(A), P(B)>0 Proof : We have DIT 111 PROBABILITY AND QUEUEING THEORY 9 NOTES Anna University Chennai P(A) = n(A) , P(B) = n(B) and P(A1B) = n(A1B) —————(1) n(S) n(S) n(S) For the conditional event A/B, favorable outcomes be one of the simple points of B, i.e, for the event A/B, the sample space is B and out of the n(B) of sample points, n(A1B) pertain to the occurrence of event A. Hence P(A/B) = n(A1B) n(B) Rewriting (1) as P(A1B) = n(B) . n(A1B) n(S) n(B) = P(B) P(A/B) Similarly we can prove P(A1B) = n(A) . n(A1B) n(S) n(A) = P(A) P(B/A) Note: Conditional probabilities P(B/A)and P(A/B) are defined if and only if P(A) | 0 and P(B) | 0 respectively. P(A/A) = 1 If A and B are independent, then P(A/B) = P(A) and P(B/A) = P(B). If A and B are independent their compliments are also independent. Theorem: If the events A and B are independent, the events Ā and B (and similarly A and B )are also independent. Proof: The events A1B and Ā1B are mutually exclusive such that (A1B) U (Ā1B) = B P (A1B) + P (Ā1B) = P (B) by addition theorem P (Ā1B) = P (B) – P (A1B) = P (B) – P(A)P(B) by product theorem = P(B)[ 1 – P(A) ] = P(B)P(Ā) Based on the above discussed theory some problems are illustrated. Most of the examples are dealing with real life experiments so that it will be interesting for you . DIT 111 PROBABILITY AND QUEUEING THEORY 10 NOTES Anna University Chennai Example 1: An integer is chosen from 2 to 15 . What is the probability that it is prime? Solution: Since there are 14 numbers from 2 to 15 the total number of numbers is 14. Out of 14 numbers 2,3,5,7,11,13 are prime numbers. Hence the number of favourable numbers is 6. Hence the probability that the number chosen is prime is No: of favorable cases = 6 = 3 Total no: of cases 14 7 Example 2: What is the chance that a) leap year selected at random will contain 53 Sundays b)a non-leap year selected at random will contain 53 Sundays? Solution: a) In a leap year, there are 366 days i.e, 52 weeks and 2 days, remaining 2 days can be any two days of the week. The different possibilities are: Sunday and Monday, Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday. In order to have 53 Sundays, out of remaining two days one must be Sunday. No: of favorable cases = 2 Total no: of cases = 7 Required Probability = 2/7 b) In a non leap year, there are 365 days i.e, 52 weeks and 1 day. The remaining 1 day can be any day of the week. Total no: of cases = 7 There will be 53 Sundays if the remaining one day is Sunday. No: of favorable cases = 1 Required probability = 1/7. Example 3: If we draw a card from a well shuffled pack of 52 cards, what is the probability that the card is either an ace or a king? Solution: The events that the card drawn is either an ace or a king are mutually exclusive. Let the event of the ace drawn be denoted by A and the king drawn be denoted by K. A pack has 4 aces and a kings. Thus P(A) = 4/52 and P(K) = 4/52 The probability that the card is either ace or a king = DIT 111 PROBABILITY AND QUEUEING THEORY 11 NOTES Anna University Chennai P(A or K) = P(A U K) = P(A) + P(K) = 4/52 + 4/52 = 2/13. Example 4: A,B and C toss a coin. The one who get the head first wins. What are their respective chances? Solution: The probability that A may win in the first round = ½. He may win in the second round, after all of them failed in the first round, with the probability = (½.½ ½). ½ = (½) 3 ½, iin the third round with probability = (½) 6 ½..and so on. Now by addition theorem the chance of A’s success = ½ + (½) 3 ½ + (½) 6 ½ + - - - - which is a G P with first term ½ and common ratio = (½) 2. Therefore the above expression is = . ½ . = 4/7. 1 - (½) 3 B may win the first round after A’s failure with probability = (½)( ½)= ¼, the second round with probability = (½) 3 ¼ and so on. Therefore B’s chances of success = ¼ +¼ (½) 3 + - - - = . ¼ . = 2/7 1 - (½) 3 Similarly, the chance of winning the game for C = . 1/8 . = 1/7 1 - (½) 3 Example 5: If from a lottery of 30 tickets numbered 1, 2, 3, - - - - 30 four tickets are drawn, what is the chance that those marked 1 and 2 are among them? Solution: Out of 30 tickets 4 tickets can be chosen in 30C 4 ways. In the 4 tickets 1 and 2 should be present. The remaining 2 tickets should be chosen from tickets numbered 3, 4, - - - -30 i.e. 28tickets which can be done in 28C 2 ways. Thus the required probability is 28C 2 = 2 30C 4 145 Example 6: A problem in mathematics is given to five students A 1 , A 2 , A 3 , A 4 and A 5 . Their chances of solving it are 1/6, 1/5, ¼, 1/3, ½ respectively. What is the probability that the problem will be solved? Solution: The probability that A 1 fails to solve the problem = 1 – 1/6 = 5/6. The probability that A 2 fails to solve the problem = 1 – 1/5 = 4/5. The probability that A 3 fails to solve the problem = 1 – 1/4 = 3/4. The probability that A 4 fails to solve the problem = 1 – 1/3 = 2/3. The probability that A 5 fails to solve the problem = 1 – 1/2 = 1/2. DIT 111 PROBABILITY AND QUEUEING THEORY 12 NOTES Anna University Chennai The probability that the problem is not solved by all five students = 5 4 3 2 1 = 1 6 5 4 3 2 6 Therefore the probability that the problem will be solved 1 – 1 = 5 6 6 Example 7: A lot consists of 10 good articles, 4 with minor defects and 2 with majoe defects. Two articles are chosen from the lot at random (without replacement). Find the probability that i) both are good, ii)both have major defects, iii) at least 1 is good, iv) at most 1 is good v)exactly 1 is good vi)neither has major defects and vii) neither is good? Solution: Since the articles are drawn without replacement, we can consider that both the articles are drawn simultaneously. i) P(both are good) = No : of ways of drawing 2 articles from good articles Total no : of ways of drawing 2 articles. = 10C 2 = 3 16C 2 8 ii) P(both have major defects) = No : of ways of drawing 2 articles with major defects Total no : of ways of drawing 2 articles. = 2C 2 = 1 16C 2 120 iii) P(atleast 1 is good) = P (exactly 1 is good and 1 is bad or both are good) = 10C 1 x 6C 1 + 10C 2 = 7 16C 2 8 iv)P(atmost 1 is good) = P(none is good or 1 is good and 1 is bad) = 10C 0 x 6C 2 + 10C 1 x 6C 1 = 5 16C 2 8 v) P(exactly 1 is good) = P(1 is good and 1 is bad) = 10C 1 x 6C 1 = ½ 16C 2 DIT 111 PROBABILITY AND QUEUEING THEORY 13 NOTES Anna University Chennai vi) P(neither has major defects) = P(both are non-major defective articles) = 14C 2 = 91 16C 2 120 vii)P(neither is good) = P(both are defective) = 6C 2 = 1 16C 2 8 Example 8: A and B throw alternately with a pair of dice. A wins if he throws (sum of numbers on top two faces) 6 before B throws 7, and B wins if he throws 7 before A throws 6. If A begins, find his chance of winning. Solution: The sum 6 can he got in 5 ways.[(1,5), (2,4),(3,3)(4,2)(5,1)] The probability of A throwing 6 is 5/36. Therefore the probability A throwing 6 = 1 – 5/36 = 31/36 Similarly, the probability of B throwing 7 is 6/36. Therefore the probability of B not throwing 7 = 1 – 6/36 = 5/6. Now A can win if he throws 6 in the 1 st , 3 rd , 5 th , - - - - -throws. Hence the chance of A winning = 5 + 31 5 5 + 31 5 31 5 5 + - - - - 36 36 6 36 36 6 36 6 36 = . 5 1 + 31.5 + 31.5 2 + 31. 5 3 + - - - - - 36 36.6 36.6 36.6 = 5 1 = 30 36 1 – 31 . 5 61 36 . 36 Example 9: A box contains 4 bad and 6 good tubes. Two are drawn out from the box at a time. One of them is tested and found to be good. What is the probability that the other one is also good? Solution: Let A= one of the tubes drawn is good and B = the other tube is good.. DIT 111 PROBABILITY AND QUEUEING THEORY 14 NOTES Anna University Chennai P(A1B) = P(both tubes drawn are good) = 6C 2 = 1 10C 2 3 With the condition that one is good, we have to find the conditional probability that the other tube is also good is required. i.e P(B/A) is required. P(B/A) = P(A1B) = 1/3 = 5 P(A) 6/10 9 Example 10: The odds against the student X solving a problem in mathematics are 8:6 and odds in favour the student Y solving the same problem are 14:16 i) What is the chance that the problem will be solved if they both try? ii) what is the probability that both working independently will solve the problem? Iii) What is the probability that neither solves the problem? Solution: Let A be the event that the student X solves the problem, and B be the evnt that the student Y solves the problem. Then by data P(Ā) = 8/14 and P(B) = 14/30. Thus, P(A) = 1 - P(Ā) = 1 – 8/14 = 6/14 P( B ) = 1 –P(B) = 1 – 14/30 = 16/30 i)Probability that the problem will be solved = P( any one solves the problem) = P(A or B) = P(A) +P(B) – P(A1B) = P(A) +P(B) – P(A)P(B) = 6/14 + 14/30 – (6/14 x 14/30) ii) Probability of solving the problem if they work independently is P(A and B) = P(A1B) = P(A)P(B) = 6/14 x14/30 iii) Probability that both will not solve the problem is = P( A and B ) = P(A)P(B) = 1/14 x 16/30 1.3.4 Theorem of total probability If B 1 ,B 2 , - - - Bn be a set of exhaustive and mutually exclusive events, and A is another associated with B i , then n P(A ) = Σ P(B i )P(A /B i ) i = 1 DIT 111 PROBABILITY AND QUEUEING THEORY 15 NOTES Anna University Chennai 1.3.5 Baye’s Theorem or Theorem of probability of causes If B 1 ,B 2 , - - - Bn be a set of exhaustive and mutually exclusive events associated with a random experiment and A is another event associated with B i , then P(B i / A) = P(B i )P(A /B i ) . , i = 1,2,3, - - n n Σ P(B i )P(A /B i ) i = 1 Proof: P(B i 1A) = P(B i ) x P(A /B i ) = P(A) x P(B i / A) P(B i / A) = P(B i ) x P(A /B i ) P(A) = P(B i ) x P(A /B i ) n Σ P(B i )P(A /B i ) i = 1 Example 1: Bolts are manufactured by three machines A, B, C. A turns out twice as many bolts as B and machines B and C produce equal number of bolts. 2% of bolts produced by A and by B are defective and 4% of bolts produced by C are defective. All bolts are put into one stock pile and one is chosen from this pile. What is the probability that it is defective? Solution: Let E 1 , E 2 , E 3 be the events that the bolts are manufactured by machines A, B,C respectively. Hint: given A turns out twice as many bolts as B and machines B and C produce equal number of bolts. P(A) = 2P(B) & P(B) = P(C) But P(A) + P(B) + P(C) = 1  2P(B) + P(B) + P(B) = 1  4P(B) = 1  P(B) = ¼  P(C) = ¼  P(A) = ½ DIT 111 PROBABILITY AND QUEUEING THEORY 16 NOTES Anna University Chennai Therefore P(E 1 ) = ½ , P(E 2 ) = ¼, P(E 3 ) = ¼ Let D be the event of chosing a defective bolt. P(D/E 1 ) = Probability of choosing a defective bolt from machine A = 2/100 P(D/E 2 ) = Probability of choosing a defective bolt from machine B = 2/100 P(D/E 3 ) = Probability of choosing a defective bolt from machine C = 4/100 Probability of choosing a defective bolt 3 P(A) = Σ P(E i ) P(A /E i ) by total probability theorem i =1 = 1 2 + 1 2 + 1 4 = 1 2 100 4 100 4 100 40 Example 2: The contents of urns I, II, and III are as follows: 2 white, 3blacks and 4 red balls; 3 white , 2 black and 2 red balls and 4 white, 1 black and 3 red balls. An urn is chosen at random and two balls are drawn. They happen to be white and red. What is the probability that they come from urns I, II, and III? Solution: Let E 1 , E 2 and E 3 denote the events that the urn I, II, and III be chosen respectively and let A be the event that the two balls taken from the urn be white and red. P(E 1 ) = P(E 2 ) = P(E 3 ) =1/3 P(A/E 1 ) = 2C 1 x 4C 1 = 2/9 9C 2 P(A/E 2 ) = 3C 1 x 2C 1 = 2/7 7C 2 P(A/E 3 ) = 4C 1 x 3C 1 = 3/7 8C 2 Now we have to calculate P(E 1 /A), P(E 2 /A) and P(E 3 /A) P(E 1 / A) = P(E 1 )P(A / E 1 ) . , 3 Σ P(E i )P(A /E i ) i = 1 DIT 111 PROBABILITY AND QUEUEING THEORY 17 NOTES Anna University Chennai = . (1/3 )(2/9) . (1/3.2/9) + (1/3.2/7) + (1/3.3/7) P(E 2 / A) = P(E 2 )P(A / E 2 ) . , 3 Σ P(E i )P(A /E i ) i = 1 = . (1/3 )(2/7) . (1/3.2/9) + (1/3.2/7) + (1/3.3/7) P(E 3 / A) = 1 – (P(E 1 / A) + P(E 2 / A)) Example 3: An urn contains5 white and 3 green balls and another urn contains 3 white and 7 green balls. Two balls are chosen at random from the first urn and put into the second urn . Then a ball is drawn from the second urn. What is the probability that it is a white ball? Solution: The two balls drawn from the first urn can be i)both white which is denoted byE 1 ii) both green which is denoted byE 2 iii)one white and one green which is denoted by E 3 P(E 1 ) = 5C 2 = 5 8C 2 14 P(E 2 ) = 3C 2 = 3 8C 2 28 P(E 3 ) = 5C 1 3C 1 = 15 8C 2 28 After the balls are transferred from the first urn to the second urn, the second urn will contain i) 5 white and 7 green balls ii) 3 white and 9 green balls iii) 4 white and 8 green balls Let A be the event of drawing a white ball from the second urn. Then DIT 111 PROBABILITY AND QUEUEING THEORY 18 NOTES Anna University Chennai P(A / E 1 ) = 5C 1 = 5 12C 1 12 P(A / E 2 ) = 3C 2 = 3 12C 2 12 P(A / E 3 ) = 4C 1 = 4 12C 2 12 Required probability of choosing a white ball = P (A ) 3 = Σ P(E i ) P(A / E i ) (by total probability theorm ) i =1 = 5 x 5 + 3 x 3 + 15 x 14 14 12 28 12 28 12 = 125 = 0.372 336 Example 4: A toy is rejected if the design is faulty or not. The probability that the design is faulty is 0.1 and that the toy is rejected if the design is faulty 0.95 and otherwise 0.45. If a toy is rejected, what is the probability that it is due to faulty design? Solution: Let D 1 , D 2 denote the events that the design is faulty or not. Let A denote the event that the toy is rejected. P(D 1 ) = 0.1 P(D 2 ) = 1 – 0.1 = 0.9 P(A/D 1 ) = 0.95 and P(A/D 2 ) = 0.45 P(of rejection due to faulty design) = P(D 1 /A) = . P(D 1 ) P(A/D 1 ) . P(D 1 ) P(A/D 1 P(D 2 ) P(A/D 2 ) = . 0.1 x 0.95 . = 0.19 0.1 x 0.95 + 0.9 x 0.45 Example 5: For a certain binary communication channel, the probability that a transmitted ‘0’ is received as a ‘0’ is 0. 95 and the probability that a transmitted ‘1’ is received as ‘1’ is 0.90. If the probability that a ‘0’ is transmitted is 0.4, find the probability that DIT 111 PROBABILITY AND QUEUEING THEORY 19 NOTES Anna University Chennai i) a ‘1’ is received and ii) a ‘1’ was transmitted given that a ‘1’ was received Solution : Let A = the event of transmitting ‘1’. Ā = the event of transmitting’0’, B = the event of receiving ‘1’ and B = the event of receiving ‘0’. Given: P ( Ā ) = 0.4, P (B/A) = 0.9 and P ( B / Ā ) = 0.95 Thus P(A) = 0.6 and P(B / Ā ) = 0.05 By the theorem of total probability P(B) = P(A) x P(B/A) + P(Ā) x P(B/ Ā) = 0.6 x 0.9 + 0.4 x 0.05 = 0.56 By Baye’s theorem P(A/B) = P(A) x P(B/A) = 0.6 x 0.9 = 27 P(B) 0.56 28 1.3.6 Bernoulli’s Trials Let us consider n independent repetitions of a random experiment. If A is an event associated with E such that P(A) remains the same for the repetitions, the trials are called Bernoulli’s trial. Theorem: If the probability of occurrence of an event in a single trial of Bernoulli’s experiment is p, then the probability that the event occurs exactly x times out of n independent trials is equal to nCr q n-x p x , where q = 1 – p, the probability of failure of the event. Example: A die is tossed until 6 appears. What is the probability that it must be tossed more than 4 times. Solution: P(X = x) = (1/6) (5/6) x – 1 x = 1,2,3, - - - - P(x>4) = 1 – P(x = 4) 4 = 1 - Σ (1/6) (5/6) x – 1 x = 1 = 1 – (1/6)[1 + 5/6 + 25/36 + 125/216] = 0.48225 Have you understood ? 1) Which of the following is an experiment? DIT 111 PROBABILITY AND QUEUEING THEORY 20 NOTES Anna University Chennai a) Tossing a coin. b) Rolling a single 6-sided die. c) Choosing a marble from a jar. d) All of the above 2) Which of the following is an outcome? a) Rolling a pair of dice. b) Landing on red. c) Choosing 2 marbles from a jar. d) None of the above. 3) Which of the following experiments does NOT have equally likely outcomes? a) Choose a number at random from 1 to 7. b) Toss a coin. c) Choose a letter at random from the word SCHOOL. d) None of the above 4) A number from 1 to 11 is chosen at random. What is the probability of choosing an odd number? a) b) c) d) None of the above. Answers: 1) d. 2)b. 3)c. d)c. Short answer questions: 1. What is random experiment? Give an example. 2. Give the axiomatic definition of probability. 3. State axioms of probability. 4. State addition theorem as applied to any 2 events .Extend it to any three events. 5. In a random experiment P(A) =1/12,P(B) = 5/12 and P(B/A) = 1/15 find P (AUB) (Solution: 89/180) 6. If P(A) = 0.5, P(B) = 0.3 and P(A1B) = 0.15, find P(A/ B ) 7. State the theorem of total probability. 8. Satate Baye’s theorem. DIT 111 PROBABILITY AND QUEUEING THEORY 21 NOTES Anna University Chennai TRY YOURSELF ! 1. From a bag containing 3 red and 2 black balls, 2 balls are drawn at random. Find the probability that they are of the same colour? (Solution:2/5) 2. Event A and B are such that P(A + B) =3/4, P(AB) = ¼ and P(Ā) = 2/3 find P(B) (Solution:2/3) 3. In a random experiment P(A ) = 1/12, P(B) = 5/12 and P(B/A) = 1/15. Find P (AUB) (Solution: 89/180) 4. If P(A) = 0.5, P(B) = 0.3 and P(A1B) = 0.15, find P(A/ B ) (Solution: 0.5) 5. Probability that India wins a cricket match against Australia is known to be 2/5 If India and Australia play 3 test 3 matches what is the probability that i) India will loose all the three matches ii) India will win all the tests iii) India will win atmost one match. (Solution: i) 27/125 ii) 98/125 iii) 8/125 iv) 81/125) 6. Two weak students attempt to write a program. Their chances of writing the program successfully is 1/8and 1/12 and the chance of making a common error is 1/10001.Find the chance that the program is correctly written. (Solution: 0.9924) 7. A fair dice is rolled 5 times. Find the probability that 1 shows twice, 3 shows twice and 6 shows once. (Solution: 0.0039) 8. One integer is chosen at random from the numbers 1,2,3,....100.what is the probability that the chosen number is divisible by (i)6 or 8and (ii) 6 or 8or both. (Solution: 1/5 , 6/25) 9. Urn I has 2 white and 3 black balls, urn II contains 4 white and 1 black balls and urn III contains 3 white and 4 black balls. An urn is selected at random and is found to be white. Find the probability that urn I was selected. (Solution: 14/57) 1.4 RANDOM VARIABLES You have seen a number of examples of sample spaces associated with various experiments. In some, the outcomes were numerical and in some others the outcomes were non numerical. DIT 111 PROBABILITY AND QUEUEING THEORY 22 NOTES Anna University Chennai For example in the experiment concerned with tossing a dice, we may have any one of the following outcome as 1,2,3,4,5 and 6 which is numerical in nature. While the result of a coin tossing experiment in which a coin is tossed once, we have the outcome and head or tail which is non numerical in nature. As it is often convenient to describe the outcome of a random experiment by a number, we will assign a number to each non numerical outcome of the experiment.. For example in the coin tossing experiment, we can assign the value of 0 to the outcome of getting heads and 1 to the outcome of getting tails. Thus in any experimental situation we can assign a real number x to every element s of the sample spaces. Random Variable – Let E be an experiment and S a sample space associated with the experiment. A function X assigning to each element s ε S, a real number X is called a random variable. The set of values which the random variable X takes is called the spectrum of the random variable. 1.4.1 Types of Random Variable – a) Discrete Random Variable b) Continuous Random Variable 1.4.1.1 Discrete Random Variable – A random variable is said to be discrete if it assumes only a finite or countably infinite values of X, that is the range space R contains a finite or countably infinite points. The possible values of x may be listed as x 1 , x 2 …. In the finite cases the list terminates. In the countably infinite cases, the list continues. Let x 1 ,x 2 …. be possible values of a discrete random variable X. Then P(x i ) is called the probability function or probability mass function or point probability function of the discrete random variable X if i) P(x i ) ¡ 0 or i=1,2…… ii) Σ P(x i ) = 1 i The collection of pairs ( x i , P(x i )), i=1,2…. is called the probability distribution. 1.4.1.2 Continuous Random Variable – A random variable X is called a continuous random variable if X takes all its possible values of an interval (or) equivalently. If the range space R x of the random variable X is an interval or a collection of intervals, X is called continuous random variable. DIT 111 PROBABILITY AND QUEUEING THEORY 23 NOTES Anna University Chennai Probability Density function (p.d.f) If x is a continuous random variable such that P{ x – dx/2 = X = x + dx/2 } = f(x)dx, then f(x) is called the probability density function ( p.d. f ) of X provided f(x) satisfies the following conditions i) f(x) ¡ 0 for all x ε R x ii) I f(x) dx = 1 R x Where R x is the range of X 1.4.1.3 Properties 1) f(x) >= 0 for all x ε R x · 2) I f(x) dx = 1 or I f(x) dx = 1 · R x b 3) P( a < x < b ) = I f(x) dx a 4) Probability at a particular point is zero. i.e it is impossible that a continuous random a variable assumes a specific value since, P(X = a) = P ( a , X , a ) = I f(x) dx =0 a This means that it is almost impossible that a continuous random variable assumes a specific value. Hence P(a ,X , b) = P ( a , X< b ) = P( a< X,b) = P( a<X<b) 1.4.2 Cumulative Distribution function (c.d.f) If X is a random variable, discrete or continuous, then P ( X,x) is called the cumulative distribution function of X or distribution fn of X and denoted a F(x) If X is discrete, If X is Continous, x F(x) = Σ Pj F(x) = I f(x) dx j · x j , x DIT 111 PROBABILITY AND QUEUEING THEORY 24 NOTES Anna University Chennai 1.4.2.1 Properties 1) F(x) is a non decreasing function of x, i.e if x 1 < x 2 then F(x 1 ) =F( x 2 ) 2) F(- ·) = 0 and F(·) = 1 3) P ( a , X , b ) can also be expressed in terms of distribution function as P ( a , X , b ) = F(b) – F(a) 4) If F(x) is a Cumulative Distribution function of a continuous random variable X then the Probability Density function of X, f (x) is given by f (x) = d [ F(x) ] dx 5) If X is a discrete random variA1Ble taking values x 1 , x 2 …. .where x 1 < x 2 < x 3 ….. <x i-1 <x i < - - - then P(X = x i ) = F(x i ) – F ( x i-1 ) Example1: The probability function of a random variable X is given by P(x) = ¼ for x = 2 = ¼ for x = 0 = ½ for x = 10. Verify that the total probability is 1.Evaluate the following probabilities a)P(X,0) b)P(X<0) c) P(|X| , 2) and d) P(0 , X ,10). Solution: · Σ p j = ¼ + ¼ + ½ = 1 j=1 Hence the total probability is 1 a) P(X,0) = P(X= -2) + P(X= 0) = ¼ + ¼ = ½ b) P(X<0) = P(X= -2) = ¼. c) P(|X| ,2) = P(-2 , X , 2) = P(X= -2) + P(X= 0) = ½. d) P(0 , X ,10) = P(X= 0) + P(X= 10) = ¾. Example 2: Consider a random experiment of tossing three times a fair coin. Let X denote the number of tails and Y denotes the number of consecutive tails. Find i)The probability distribution of X and Y ii)the distribution function of x iii)the probability distribution of X + Y and XY. DIT 111 PROBABILITY AND QUEUEING THEORY 25 NOTES Anna University Chennai Solution: The sample space of the random experiment is S= {HHH, HHT, HTH,THT,HTT,THH,TTH,TTT} Each element of S occurs with probability 1/8. The values of X, Y, X + Y and XY for each outcome is tabulated. Event HHH HHT HTH HTT THH THT TTH TTT X 0 1 1 2 1 2 2 3 Y 0 0 0 2 0 0 2 3 X+Y 0 1 1 4 1 2 4 6 XY 0 0 0 4 0 0 4 9 i)Probability distribution of X Value of X, x 0 1 2 3 p(x) 1/8 3/8 3/8 1/8 ii)Probability distribution of Y Value of Y, y 0 2 3 P(y) 5/8 2/8 1/8 iii)distribution function of X X [-·,0) [0,1) [1,2) [2,3) [3,·) F(X) 0 1/8 4/8 7/8 1 DIT 111 PROBABILITY AND QUEUEING THEORY 26 NOTES Anna University Chennai iv)Probability distribution of X + Y Value of X+Y 0 1 2 4 6 p(x) 1/8 3/8 1/8 2/8 1/8 v) Probability distribution of XY Value of XY 2 4 9 p(x) 5/8 2/8 1/8 Example 3: Find the constant c so that the function f(x) = cx ,2 , 0<x<3 = 0, otherwise is a pdf . Find the distribution function and evaluate P(1<x<2). Solution: Given f(x) = cx ,2 , 0<x<3 = 0, otherwise By the property of pdf · I f(x)dx = 1 - · 3 3 I cx ,2 dx = 1 => [cx 3 /3] = 1 => 9c =1 => c = 1/9. 0 0 The distribution function is given by x F(x) = P(X , x) = I f(x)dx - · x When x , 0, F(x) = P(X , 0) = I f(x)dx = 0 since f(x) is not defined for x<0. - · 0 x x x When 0<x<3, F(x) = P(X,x) = I f(x)dx +I f(x)dx = I x 2 /9 dx = x 3 /27] = x 3 /27. - · 0 0 0 When x=3 0 3 x 3 3 F(x) = P(X ,x) = I f(x)dx + I f(x)dx + I f(x)dx = I x 2 /9 dx = x 3 /27] = 1 - · 0 3 0 0 DIT 111 PROBABILITY AND QUEUEING THEORY 27 NOTES Anna University Chennai F(x) = 0, x=0 x 3 /27, 0<x<3 1, x =3. 2 2 2 P(1<x<2) = I f(x)dx = I x 2 /9dx = 1/9(x 3 /3) = 7/27. 1 1 1 Example 4: A random variable has the following probability distribution X : 0 1 2 p(x) : 3c 2 4c – 10c 2 5c - 1 Find i)the value of c ii) P(0<X<2/X>0) and iii) the distribution function of X iv)the largest value of X for which F(x) < ½ and v) smallest value of X for which F(x) > ½ Solution: i)since Σp(x) = 1, 3c 2 + 4c – 10c 2 + 5c -1 = 1 7c 2 -9c + 2 = 0 c= 2/7, 1 The value c = 1makes some p(x) negative which is meaningless. Therefore c = 2/7 ii) P(0<x<2/x>0) We know P(A/B) = P(A1B) P(B) P(0<x<2/x>0) = P(0<x<2 1 x>0) = 4c – 10c 2 = 4c – 10c 2 where c =2/7 P(x>0) P(x=1) + P(x=2) 4c – 10c 2 + 5c-1 = 4c – 10c 2 = 8/7 - 40/49 = 16/37 – 10c 2 + 9c – 1 -40/9 +18/7 -1 iii)cdf is defined as F(x) = P(X=x) when x<0 F(x) = 0 when 0 , x<1 F(x) = P(X = 0) = 12/49. when 1 , x<2 F(x) = P(X = 0) + P(X = 1) = 4/7 when x ¡ 2 F(x) = P(X = 0) + P(X = 1) + P(X = 2) = 1 iv) form the above cdf it is clear that the largest value of X for which F(x)<½ isx = 0 v) Similarly form the above cdf it is clear that the smallest value of X for which F(x) > ½ is x = 1 DIT 111 PROBABILITY AND QUEUEING THEORY 28 NOTES Anna University Chennai Example 5: The probability function of an infinite discrete distribution is given by P(X=j) = 1/2 j (j = 1,2,…8). Verify that the total probability is 1 and find the mean and variance of the distribution. Find also P(X is even). P(X=5) and P(X is divisible by 3). Solution: Let P(X =j) = p j · 3 p j = ½ +1/2 2 1/2 3 + - - - - - · which is geometric series j =1 = ½ = 1 1- ½ · The mean of X is defined as E(X) = 3 j p j j =1 Therefore E(X) = a +2a 2 +3a 3 + - - - - · where a = 1/2 = a(1 + 2a + 3a 2 + - - - - · ) = a(1 - a) -2 = ½ = 2. (½) 2 The variance of X is defined as V(X) = E(X 2 ) – [E(X)] 2 · Where [E(X)] 2 = 3 j 2 p j j=1 · [E(X)] 2 = 3 j 2 a j , where a = 1/2 j =1 · · · = 3 [j(j + 1) – j] a j = 3 j(j + 1) a j - 3 ja j j=1 j =1 j =1 = a(1.2+2.3a+3.4a 2 + - - - - · ) –a(1 + 2a +3a 3 + - - - · ) = a x 2(1-a) 3 – a x (1-a) -2 = 2a - a (where a =1/2) (1-a) 3 (1-a) 2 = 8-2 = 6 Variance = E(X 2 ) – [E(X)] 2 = 6 - 4 = 2 DIT 111 PROBABILITY AND QUEUEING THEORY 29 NOTES Anna University Chennai P(X is even) = P(X = 2 or X = 4 or X= 6 or- - - - - - ) = P(X = 2 )+P(X = 4) + P( X= 6) +- - - - - - · =(½) 2 +(½) 4 +(½) 6 + +- - - - - - · = ¼ 1 - ¼ = 1/3 P(X=5) = P(X = 5 or X = 6 or X= 7 or- - - - - - ) = P(X = 5 )+P(X = 6) + P( X= 7) +- - - - - - · = (½) 5 1- ½ = 1/16 P(X is divisible by 3) = P(X = 3 or X = 6 or X= 9 or- - - - - - ) = P(X = 3 )+P(X = 6) + P( X= 9) +- - - - - - · = (½) 3 +(½) 6 +(½) 9 + +- - - - - - · = 1/8 1 – 1/8 = 1/7 Example 6: The diameter of an electric cable X is a continuous RV with pdf f(x) = kx(1 -x), 0=X=1. Find i)the value of k, ii)cdf of X, iii) the value of a such that P(X<a) = 2P(X>a) and iv) P(X , ½/1/3<X<2/3) Solution i) If f(x) is pdf I f(x)dx = 1 1 1 1 I kx(1 -x)dx = 1 => I k(x –x 2 )dx = 1 => k[x 2 /2 - x 3 /3] = 1 0 0 0 => k/6 =1 => k =6 x ii) F(x) = P(Xsx) = I f(x)dx - · x When xs0, F(x) = P(Xs0) = I f(x)dx = 0 since f(x) is not defined for x<0. - · DIT 111 PROBABILITY AND QUEUEING THEORY 30 NOTES Anna University Chennai 0 x x x When 0sxs1, F(x) = P(Xsx) = I f(x)dx +I f(x)dx =Ik(x –x 2 ) dx =k[x 2 /2- x 3 /3 ] =. - · 0 0 0 = 3x 2 – 2x 3 When xs1 0 1 x 1 1 F(x) = P(Xsx) = I f(x)dx + I f(x)dx + I f(x)dx = I k(x –x 2 ) dx = k[x 2 /2 - x 3 /3 ] - · 0 1 0 0 = 1 Therefore cdf is given by F(x) = 0, x , 0 3x 2 – 2x 3 , 0 , x , 1 1, x ¡1. iii) P(X<a) = 2P(X>a) a · => If(x)dx = 2 I f(x)dx 0 a a 1 => If(x)dx = 2 I f(x)dx 0 a a 1 => I k(x –x 2 )dx = 2 I k(x –x 2 )dx 0 a a 1 => k[x 2 /2 - x 3 /3 ] = 2 k[x 2 /2 - x 3 /3 ] 0 a => a 2 /2 - a 3 /3 = 2[1/2 – 1/3 - a 2 /2 + a 3 /3 ] => 3a 2 /2 - 3a 3 /3 - 1/3 = 0 => 9a 2 – 6a 3 -2 = 0 6 The value of a is the solution of the cubic equation 6a 3 -9a 2 + 2 = 0, which lies between o and 1. iv) P(0, X, 1/2 n 1/3 < X < 2/3) = P(1 /3 < X < 1/2) P(1 /3 < X < 2/3) P(1 /3 < X < 2/3) 0 1/3 ½ 2/3 1 DIT 111 PROBABILITY AND QUEUEING THEORY 31 NOTES Anna University Chennai 1/ 2 1/2  I k(x –x 2 ) dx [x 2 /2 - x 3 /3 ] 1/8– 1/24 - 1/18 + 1/81 1/3 = 1/3 = ½ 2/3 I k(x –x 2 ) dx [x 2 /2 - x 3 /3 ] 4/18 - 8/81 -1/18 +1/81 1/3 1/3 = (13/24) / (13/162) = ½ Example 7: Suppose that the life of a certain radio tube (in hours) is a continuous R V with pdf f(x) = 100/x 2 , x ¡100 = 0, elsewhere i) Find the distribution function ii) If 3 such tubes are inserted in a set what is the probability that none of the three tubs will be replaced during the first 150 hours of operation. iii) What is the probability that exactly one tube will have to be replaced after 150 hours of service? iv) What is the probability that all the tubes will be replaced after 150 hours of service? v) What is the maximum number of tubes that may be inserted into a set so that there is probability of 0.1 that after 150 hour of service all of them are still functioning? Solution: The distribution function is given by P(X ,x) When x , 100 F(x) = P(X , x) = 0 since f(x) is 0 When x>100 x x x F(x) = P(X,x) = I f(x)dx = I 100/x 2 dx = 100 [-1/x] = 1 - 100/x 100 100 100 The distribution function is F(x) = 0 , x , 100 = 1 - 100/x , x > 100 ii) Let the random variable X be the life of a radio tube. Probability of a tube will last for 150 hours = P(X , 150) 150 = I (100/x 2 ) dx = 1/3. 100 DIT 111 PROBABILITY AND QUEUEING THEORY 32 NOTES Anna University Chennai Therefore the probability that one tube will not be replaced in the first 150 hours is 1/3. The probability of two will not be replaced in the first 150 hours is (1/3) 2 . Probability that none of the three tubes will have to replaced during first 150 hours is (1/3) 3 iii)The probability that a tube will not last for 150 hours = 1-1/3 = 2/3. Probability that two tubes last for 150 hours = (1/3) 2 .Since there are 3C 1 ways of choosing 1 tube from the 3 tubes, the required probability that exactly after 150 hours = 3C 1 (1/3) 2 (2/3) = 2/9. iv) Probability that 1 tube will be replaced after 150 hours = 2/3 Thus the probability that all the tubes will be replaced after 150 hours = (2/3) 3 = 8/27. v)Suppose there are n tubes. The probability that all the n tub4es are functioning after 150 hours is(2/3) n . Since this probability is given t be 0.1, we have (2/3) n = 0.1. if we substitute values for n we can see that n = 5 is the maximum value of n that satisfies the equation. Hence 5 tubes are to be inserted so that all of them are functioning after 150 hours. Example 8: The c.d.f of a continuous R.V X is given by F(x) = 0, x < 0 = x 2 , 0 = x < ½ = 1 - 3/25(3-x) 2 , ½ , x < 3 Find the pdf of X and evaluate P(|X| = 1) and P(1/3 = X < 4) using both the pdf and cdf. Solution: The points x = 0,1/2 and 3 are points of continuity we know that if X is a continuous random variable d F(x) = f(x) dx therefore f(x) = 0, x<0 = 2x, 0 , x<½ =6/25(3 - x), ½ , x<3 = 0, x ¡3. Although the points x = 1/2 , 3 are points of discontinuity for f(x), we may assume that f(1/2) = 3/5 and f(3) = 0. P(|X| ,1) = P(-1, x ,1) 1 ½ 1 = I f(x)dx = I 2x dx + I 6/25(3 - x)dx = 13/25 -1 0 ½ If we use property of cdf P(|X| ,1) = P(-1 , x ,1) = F(1) - F(-1) = 13/25 DIT 111 PROBABILITY AND QUEUEING THEORY 33 NOTES Anna University Chennai If we use property of pdf ½ 3 P(1/3 , X<4) = I 2x dx + I 6/25(3 - x)dx = 8/9 1/3 ½ If we use property of cdf P(1/3 , X<4) = F(4) - F(1/3) = 1 – 1/9 = 8/9. Have you understood ? Say true or false. Justify your answer. 1. A random variable is a multi valued function. 2. Probability distribution function and probability density function of a continuous random variable are continuous. 3. For a discrete random variable, the probability density function represents the probability mass function. 4. The probability mass function can take negative values. 5. A discrete random variable can be considered as a limiting case of continuous random variable with impulse distribution. Answers:(1.False, 2. True, 3.True, 4. False, 5.True) Short answer questions. 1. Define a random variable? 2. What is discrete random variable? Give an example. 3. What is continuous random variable? Give an example. 4. What is a probability distribution function? 5. Give the properties of probability distribution function? 6. What is a probability density function? 7. Give the properties of probability density function? TRY YOURSELF ! 1. A continuous random variable X has probability density function given by f(x) = 3x 2 0, x ,1. Find K such that P(X >K) = 0.05. (Solution: k = 0.7937). 2. If the density function of a continuous RV X is given by f(X) = ax , 0 , x ,1 = a, 1, x ,2 = 3a – ax, 2 , x ,3 = 0, elsewhere i) Find the value of a ii) find the cdf of X DIT 111 PROBABILITY AND QUEUEING THEORY 34 NOTES Anna University Chennai iii)If x 1 , x 2 and x 3 are 3 independent observations of X, what is the probability that exactly one of these 3 is greater than 1.5.(Solution: i)a = ½ ii) x 4 /4, 0, x ,1; x/2 – ¼, 1, x ,2 ; 3x/2 – x 2 /4 – 5/4, 2 , x ,3; 1 x >3. iii) 3/8) 3. A random variable X has the following probability function X : 0 1 2 3 4 P(X): K 3K 5K 7K 9K i) Find the value of K ii) Find P(X<3), P(X ¡ 3), P(0<X<4) iii) Find the distribution function of X. (Solution: i) K = 1/25 ii)9/25, 16/25, 3/5, iii) 1/25,4/25,9/2516/25,1) 1.5 EXPECTATION AND MOMENTS 1.5.1 Expectation If X is a discrete random variable, then the expected value or the mean value of g(x) defined as E (g(x)) = Σ g(x i ) p i i Where, p i = P(X = x i ) is the probability mass function of X. If X is a continuous random variable with pdf. f(x) then E (g(x)) = Ig(x) f(x) dx Rx Mean µ x = E(x) = Σ x i p i if X is discrete i = I x f(x) dx if X is continuous Rx Var(x) = σ x 2 = E ((x- µ x ) 2 ) = Σ (x i - µ x ) 2 p i , if X is a discrete = I (x- µ x ) 2 f(x)dx , if X is continuous The square root of variance is called Standard Deviation. DIT 111 PROBABILITY AND QUEUEING THEORY 35 NOTES Anna University Chennai Note: E(a) = a E (aX) = a E (X) E (aX + b) = a E (X) + b Var (X) = E(X 2 ) – (E(X)) 2 Var (aX) = a 2 Var(X) Var (aX + bY) = a 2 Var(X) + b 2 Var(Y) 1.5.2 Moments: If X is a discrete or continuous random variable the n th moment about the origin is defined as the expected value of the n th power of X and is denoted by µ n ’ µ n 1 = E (X n ) = Σ x i n p i · = I x n f(x) dx - · The n th central moment of a random variable X is its moment about its mean value X and is defined as E[(X - X) n ] = Σ (x i – x ) n p i = µ n , if X is discrete · = I (x- x ) n f(x) dx = µ n , if X is continuous -· Since the first and second moments about the origin are given by µ 1 ' = E (X) and µ 2 ' = E (X 2 ) we have mean = first moment about the origin Var(X) = µ 2 ' – (µ 1 ') 2 Note: E(|X| n ) and E ( |X - µ x | n ) are called absolute moments of X. E { | X - a| n } and E {(X - a) n } are called generalized moments of X. 1.6 MOMENT GENERATING FUNCTION: The moment generating function (m g f) of a random variable X about the origin whose probability distribution function f(x) is given by E(e tX ) and is denoted by M X (t) Hence M X (t) = E(e tX ) DIT 111 PROBABILITY AND QUEUEING THEORY 36 NOTES Anna University Chennai = Σ e tx f(x), if X is a discrete · = I e tx f(x) dx , if X is continuous -· Moment generating function will exist only if the sum or integral of above definition converges. Moment generating function generates the moments µ n ’ about origin. Consider M X (t) = E(e tX ) = E 1 + tx + t 2 X 2 + - - - - + t n X n + - - - - 2! n! = 1 + tE(x) + t 2 E(X 2 ) + - - - - + t n E(X n ) + - - - - 2! n! = 1 + t µ 1 ' + t 2 µ 2 ' + - - - - + t n µ n ‚ + - - - - 2! n! where µ n 1 is the n th moment about the origin. Thus we see that the coefficient of t n / n! in M x (t) gives µ n 1 . µ 1 ' = d M X (t) dt t = 0 µ 2 ' = d 2 M X (t) dt 2 t = 0 µ n ’ = d n M X (t) dt n t = 0 Theorem: M cX (t) =M X (ct) where c is a constant. Proof: By definition of m g f M cX (t) = E(e tcX ) = E(e ctX ) = M X (ct) Theorem: The m g f of the sum of independent random variables is equal to the product of their respective m g fs. i.e If X 1 , X 2 , - - - X n are n independent random variables then the m g f of X 1 + X 2 + - - - + X n is given by M X1+ X2 + - - - + Xn (t) = M X1 (t)M X2 (t) - - -M Xn (t). Proof: By the definition of ,m g f M X1+ X2 + - - - + Xn (t) = E [e t (X1+ X2 + - - - + Xn ) ] = E [e tX1 e t X2 - - - e t Xn ) ] DIT 111 PROBABILITY AND QUEUEING THEORY 37 NOTES Anna University Chennai = E [e tX1 ][ e t X2 ] - - - [e t Xn ) ] = M X1 (t)M X2 (t) - - -M Xn (t). Theorem: (Effect of change of origin and scale on m g f): Let X be transformed to a new variable U by changing both the origin and scale in X as U = (X - a)/ h where a and h are constants .Then m g f (about the origin) of U is given by M U (t) = e -at/h M X (t/h) Proof: M U (t) = E[e tU ] = E[e t(X - a)/h ] = E[e t(X/h e -at/h ] = e -at/h E[e t(X/h ] = e -at/h M X (t/h) Note: in the above theorem putting a = E(X) = µ and h = σ (standard deviation), U = [X – E(X)] / S.D(X) = (X – μ) / σ = Z is called the standard variate. The m g f of Z is M Z (t) = e -µt/h M X (t / σ) Theorem: Uniqueness theorem If two random variables have the same moment generating function then they must have the same distribution. Example1: Find the m g f for f(x) = 2/3 at x = 1 1/3 at x = 2 0 otherwise And also find µ 1 ' & µ 2 ' Solution: M X (t) = Σ e tx f(x) = e t (2/3) + e 2t (1/3) Mean = µ 1 ' = d M X (t) = d e t (2/3) + e 2t (1/3) dt t = 0 dt t = 0 = e t (2/3) + e 2t (2/3) = 4/3 t = 0 µ 2 ' = d 2 M X (t) = d 2 e t (2/3) + e 2t (2/3) = 2 dt 2 t = 0 dt 2 t = 0 DIT 111 PROBABILITY AND QUEUEING THEORY 38 NOTES Anna University Chennai Example 2: A r.v is uniformly distributed in the interval (a,b)with p d f f(x) = 1/(b - a) , a< x<b = 0 elsewhere Find the m g f of X . Using this find the mean and variance in two different ways . Solution: By definition · M X (t) = E(e tX ) = I e tx f(x) dx , -· b = I e tx (1/b-a) dx , = e tb - e ta a t (b- a) M X (t) = 1 1 + tb + t 2 b 2 + t 3 b 3 + . . . . - 1 + ta + t 2 a 2 + t 3 a 3 + - - t(b - a) 2! 3! 2! 3! = 1 t (b - a) + t 2 (b 2 – a 2 ) + t 3 (b 3 – a 3 ) + - - - - t (b - a) 2! 3! M X (t) = 1 + t (b + a) + t 2 (b 2 + ab + a 2 ) + - - - 2! 3! Mean = µ 1 ' = d M X (t) dt t = 0 = d 1 + t (b + a) + t 2 (b 2 + ab + a 2 ) + - - - dt 2! 3! t = 0 = 0 + (b + a) + remaining terms will contain t in the numerator 2 Mean = E(X) = (b+ a) / 2 E(X 2 ) = µ 2 ' = d 2 M X (t) dt 2 t = 0 = d 2 1 + t (b + a) + t 2 (b 2 + ab + a 2 ) + - - - dt 2 2! 3! t = 0 = 2 (b 2 + ab + a 2 ) 3! DIT 111 PROBABILITY AND QUEUEING THEORY 39 NOTES Anna University Chennai Var(X) = E(X 2 ) – [E(X)] 2 = 2 (b 2 + ab + a 2 ) - [(b+ a) / 2] 2 3! = (b - a) 2 12 Another method to find mean and variance: As you know the coefficient of t n / n! in the expansion of M x (t) gives µ n 1 . Mean = µ 1 1 th coefficient of t /1! or t We have M X (t) = 1 + t (b + a) + t 2 (b 2 + ab + a 2 ) + - - - 2! 3! = 1 + t (b + a) + t 2 (b 2 + ab + a 2 ) + - - - 2 2! .3 µ 1 1 = (b + a) 2 µ 2 ‘ = (b 2 + ab + a 2 ) 3 Variance = µ 2 ‘ – (µ 1 ‘) 2 = (b - a) 2 12 Example 3: Show that m g f of the random variable X having p d f f(x) = 1/3, -1<x<2 = 0, elsewhere is given by M X (t) = e 2t – e -t , t | 0 3t = 1 , t = 0 Solution: · 2 2 M X (t) = I e tx f(x)dx = 1/3 I e tx dx =1/3 [e tx /t] = 1/3[e 2t /t – e -t /t] = e 2t - e -t , t | 0 -· -1 -1 3t When t = 0 M X (0) = 0 0 Applying L’Hospital rule M X (0) = 2e 2t + e -t = 1 3 t = 0 DIT 111 PROBABILITY AND QUEUEING THEORY 40 NOTES Anna University Chennai Therefore mg f of the random variable X is given by M X (t) = e 2t – e -t , t | 0 3t = 1 , t = 0 Example 4: Lt X be the random variable which assumes the value x with the probability P(X = x) = q x-1 p, x = 1,2,- —. Find the m g f of X and find its mean and variance. Solution: · · · M X (t) = Σ e tx f(x) = 3 e tx q x-1 p = p 3 (qe t ) x = p (qe t ) 3 (qe t ) x-1 x = 1 q x =1 q x =1 = pe t (1 - qe t ) -1 = pe t 1 - qe t M X (t) = pe t 1 - qe t Mean = µ 1 ' = d M X (t) = d pe t dt t = 0 dt 1 - qe t t = 0 = (1- qe t ) pe t - pe t (-qe t ) = (1 - q) p - p(-q) = p = p = 1 (p+ q = 1) (1 - qe t ) 2 t = 0 (1 - q) 2 (1 - q) 2 p 2 p Var(X) = µ 2 ' – (µ 1 ') 2 µ 2 ' = d 2 M X (t) = d 2 pe t = d (1- qe t ) pe t + pqe 2t ) dt 2 t = 0 dt 2 1 - qe t t = 0 dt (1 - qe t ) 2 t=0 = (1-qe t ) 2 [(1- qe t ) pe t + pe t (-qe t )+2pqe 2t ]–{[(1-qe t ) pe t + pqe 2t )]2(1-qe t ) (-qe t )} (1 - qe t ) 4 t=0 = (1 - q) 2 [(1- q) p + p(-q) + 2pq] – {[(1- q) p + pq)]2(1 - q) (-q)} (1 - q) 4 µ 2 ' = (1 - q) 2 p + 2pq(1 - q) = (1 - q)p + 2pq = p(1 + q) = 1+q (1 - q) 4 (1 - q) 3 p 3 p 2 Var(X) = µ 2 ' – (µ 1 ') 2 = 1/p - (1+q)/p 2 = q p 2 DIT 111 PROBABILITY AND QUEUEING THEORY 41 NOTES Anna University Chennai Example 5: Let X have the probability mass function f(x) = 6 1 , k = 1, 2,- - - π 2 k 2 0 otherwise. Find the m g f ? Solution: · M X (t) = Σ e tx f(x) = Σ e tk 6 1 (as k  8, e tk  8) k = 1 π 2 k 2 M X (t) is infinity for any t>0 Therefore m g f does not exist. We will be discussing about the m g f of different distributions in the coming topics. Have you understood ? 1. Define expectation of a random variable. 2.Define nth central moment about its origin. 3.Define mgf of a random variable? 4.State the properties of mgf of a random variable. TRY YOURSELF ! 1. Find the mgf for the distribution where f(x) = 2/3 at x = 1 = 1/3 at x = 2 = 0, otherwise Also find the mean and variance? (Solution: e t 2/3 + e 2t 1/3, 4/3, 2) 2. A random variable X has density function given by f(x) = 2 e -2x , x ¡ 0 = 0, x < 0 Obtain the mgf and the mean? (Solution: 2/(2 - t), ½ ) While constructing probabilistic models for observable phenomena, certain probability distributions arise more frequently than do others. We treat such distributions that play important roles as special probability distributions. Here we will be discussing discrete as well as continuous distributions in considerable detail. DIT 111 PROBABILITY AND QUEUEING THEORY 42 NOTES Anna University Chennai 1.7 DISCRETE DISTRIBUTIONS 1.7.1 Binomial Distribution Many random processes can be mapped into two outcomes. For example: - win or lose - success or failure - heads or tails These either-or situations are called binary outcomes. Suppose I am at a party, and I ask a girl to dance. There are two outcomes. She can agree to dance with me, or she can turn me down. At the party, I can ask any number of girls to dance with me. Each girl has the same two choices: agree to dance with me, or turn me down. Assume that each girl’s decision is independent of one another, and that my probability of success with each girl is the same. These are the assumptions that we described at the end of the last lecture. Let n be the number of girls I ask to dance. Let X be the number of girls who agree to dance with me. We say that X is a binomial random variable, because it is the sum of binary outcomes. Suppose that when I ask a girl to dance with me, the probability that she will agree to do so is .15. We call this p, the probability of success. The distribution of the random variable X—my overall number of dance partners— wi ll depend on the number of girls I ask, n, and my probability of success, p. It is a distribution associated with repetition of independent trial of an experiment. Each trial can result in a success with probability ‘p’ and a failure with probability q = 1-p. Such a trial is know as a Bernoulli trial. Some examples of Bernoulli trials are – 1) Toss of a single coin ( head or tail ) 2) Performance of a student in an examination ( pass or fail ) A random variable X which takes only 2 values 0 and 1 with probability q and p respectively ie P ( X = 1 ) = p, P (x = 0 ) = q. Then q = 1 – p is called a Bernoulli variable and X is said to have Bernoulli distribution. P is called the parameter of Bernoulli variate. A Bernoulli random variable with parameter rp has probability mass function given by P ( X –x ) = p x q 1 – x where x = 0,1 …. DIT 111 PROBABILITY AND QUEUEING THEORY 43 NOTES Anna University Chennai An experiment consisting of repeated number of Bernoulli trails is called Binomial experiment. A binomial experiment must posses the following properties 1) there must be a fixed number of trials 2) all trails must have identical probabilities of success 3) the trials must be independent of each other 1.7.1.1 Binomial distribution Let X be the number of success in a repeated independent Bernoulli trials with probability p of success for each trial .Then X is called the Binomial random variable with parameters p and n or symbolically B ( n, p) The probability mass function of a binomial random variable given by P ( X= x ) = n Cx p x q n - x , where x = 0,1,2 … n where p + q = 1 Note: 1) The name binomial distribution is given since the probabilities nC x q n-x p x ( x = 0,1,2….n ) are the successive terms in the expansion of the binomial expression ( p + q ) n 2) Binomial distribution is a legitimate probability distribution since n n Σ P (X = x ) = Σ nC x q n-x p x x=0 x=0 3) If we assume that n trials constitute a set and if we consider N sets, the frequency function of the binomial distribution is given by f(x) = Np(x) = N nCx r p x q n-x : x = 0,1,2, - - -n In many scientific and engineering applications the Binomial distribution finds application. 1.7.1.2 Additive property of a Binomial distribution If X 1 and X 2 are 2 input binomial random variable with parameter ( p 1 , n 1 ) and (p 1 , n 2 ) then X 1 + X 2 is a Binomial random variable with parameters ( p, n 1 + n 2 ) In many scientific and engineering applications the Binomial distribution finds application. 1) it is used in quality control statistics to count the number of defects of an item 2) in biology to count the number of bacteria 3) in insurance problems to count the number of causalities DIT 111 PROBABILITY AND QUEUEING THEORY 44 NOTES Anna University Chennai 1.7.1.3 Mean and variance of the binomial distribution E(X) = Σ x i p i n = Σ x. nC x p x q n-x x = 0 n = Σ x. n! p x q n-x (1) x = 0 x!(n - x)! n = np Σ (n-1)! p x-1 q (n-1) – (x – 1) x = 0 (x-1)!{(n - 1) – (x - 1)}! n = np Σ (n - 1)C x – 1. p x-1 q (n - 1) – (x - 1) x = 1 = (q + p) n-1 = np. (2) n E(X 2 ) = Σ x i 2 p i = Σ x 2 p x 0 n = Σ {x(x-1) + x} n! p x q n-x x = 0 x! (n - x)! n n = Σ x(x-1) n! p x q n-x + Σ x n! p x q n-r x = 0 x! (n - x)! x = 0 x! (n - x)! n = n(n - 1)p 2 Σ (n - 2)! p x-2 q (n - 2) – (x - 2) + np x = 2 (x - 2)!{(n-2) – (x - 2)}! n = n(n - 1)p 2 Σ (n - 2)C x-2 p x-2 q n-x + np x = 2 = n(n - 1)p 2 (q + p) n-2 + np = n(n - 1)p 2 + np Var (X) = E(X 2 ) + { E(X)} 2 = n(n - 1)p 2 + np - n 2 p 2 = np(1 - p) = npq DIT 111 PROBABILITY AND QUEUEING THEORY 45 NOTES Anna University Chennai 1.7.1.4 M G F of the binomial distribution: n Mx(t) = Σ e tx p x x = 0 n = Σ e tx n C x p x q n-x x = 0 n = Σ n Cx r (pe t ) x q n-x x = 0 = (pe t +q) n 1.7.1.5 Recurrence formula for the central Moments of the binomial distribution. By definition , the k th order central moment µ k = E{X – E(X)} k . For the binomial distribution B(n,p) n µ k = Σ (x - np) k n C x p x q n-x (1) x = 0 By differentiating (1) with respect to p, n dµ k = Σ nC x [-nk(x - np) k -1 .p x q n-x + (x - np) k {xp x-1 q n-x + (n - x)p x q n-x-1 (-1)}] dp x = 0 n = -nµ k-1 + Σ nC x (x - np) k p x-1 q n-x-1 [xq – (n - x)p] x = 0 n = -nµ k-1 + Σ nC x (r - np) k p r-1 q n-x-1 (x - np) (since p +q = 1) x = 0 n = -nµ k-1 + 1 Σ nC x p x q n-x (x - np) k+1 q x = 0 = -nµ k-1 + 1 µ k+1 pq i.e., µ k+1 = pq dµ k + nµ k-1 (2) dp using recurrence relation (2) we may compute moments of higher order, provided moments of lower order are known Putting k = 1 in (2), we get DIT 111 PROBABILITY AND QUEUEING THEORY 46 NOTES Anna University Chennai µ 2 = pq dµ 1 + nµ 0 dp = npq (since the values µ 0 = 1 and µ 1 = 0) Putting k = 2 in (2), we get µ 3 = pq dµ 2 + 2nµ 1 dp = pq d [np(1 - p)] dp = npq[1 – 2p] = npq(q – p) Putting k = 2 in (2), we get µ 4 = pq dµ 3 + 3nµ 2 dp = npq d {p(1-p) (1- 2p)} + 3npq dp = npq [1- 6p + 6p 2 + 3npq ] = npq [1 – 6pq + 3npq] = npq [1 + 3pq(n-2)] Note: µ 2 is the variance, µ 3 is a measure of skewness and µ 4 is a measure of kurtosis. 1.7.1.6 Recurrence formula for Binomial distribution distribution: P(X = x) = nC x p x q n-x P(X = x + 1) = nC x+1 p x+1 q n-(x+1) P(X = x + 1) = nC x+1 p x+1 q n-(x+1) P(X = x) nC x p x q n-x P(X = x + 1) = (n - x). p . P(X = x) ( x +1) . q Example 1: For a binomial distribution with parameter n = 5 and p, the probability of success = 0.3, find the probabilities of getting i) at least 3 success ii) at most 3 success iii)exactly 3 failures. DIT 111 PROBABILITY AND QUEUEING THEORY 47 NOTES Anna University Chennai Solution: Let X denote the success. Probability distribution is given nCxp x q n-x Given n = 5. p= 0.3, q = 1 – p = 1 – 0.3 = 0.7 i)the probability of atleast 3 successes = P(X = 3) + P(X = 4) + P(X = 5) = 5C 3 (0.3) 3 (0.7) 2 + 5C 4 (0.3) 4 (0.7) 1 + 5C 5 (0.3) 5 (0.7)0 = 0 .1631 ii)The probability of atmost 3 successes = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = (0.7) 5 + 5C 1 (0.3) 1 (0.7) 4 + 5C 2 (0.3) 2 (0.7) 3 + 5C 3 (0.3) 3 (0.7) 2 = .9692 iii)The probability of exactly 3 failures = the probability of exactly 2 successes = P(X = 2) = 5C 2 (0.3) 2 (0.7) 3 = .3807 Example 2 : The mean of a binomial distribution is 5 and standard deviation is 2. Determine the distribution. Solution: Mean of the Binomial distribution is 5 i.e, np = 5 S D is 2. Therefore \vnpq = 2 or npq = 4 npq = 4 np 5 q = 4/5, p = 1 – q = 1 – 4/5 = 1/5 We have np = 5, but p = 1/5 Therefore n = 25 Hence the Binomial distribution is nCxp x q n-x = 25Cx(1/5) x (4/5) n-x , x = 0,1,2, - - -25 Example 3: The mean and variance of a Binomial variate are 8and 6.Find P(X=2). Solution: Given np = 8and npq = 6. npq = 6 = 3 np 8 4 Therefore q = ¾, hence p = 1 – ¾ = ¼ But np = 8 i.e, n ¼ = 8 => n = 32 The Binomial distribution is nCxp x q n-x = 32Cx(¼) x ( ¾) 32 – x , x = 0,1,2, - - -32 DIT 111 PROBABILITY AND QUEUEING THEORY 48 NOTES Anna University Chennai P(X ¡ 2) = 1 – [p(0) + p(1)] = 1 – (¾) 32 + 32C 1 (¼)( ¾) 31 = 1 - (¾) 31 [ ¾ +3 2/4] = 1 – (35/4)( ¾) 31 Example 4: If on an average one vessel in every ten is wrecked, find the probability that out of five vessels expected to arrive, at least four will arrive safely. Solution: Let p denote the probability that a vessel will arrive safely Then p = 9/10 and q = 1/ 10 Therefore probability for at least 4 out of 5 vessels to arrive safely = P( X= 4) + P(X = 5) = 5C 4 (9/10) 4 (1/10) + (9/10) 5 = 0.9185 Example 5: In a Binomial distribution consisting of 5 independent trials, the probabilities of 1 and 2 successes are 0.4096 and 0.2048respectively. Find the parameter p of the distribution. Solution: The Binomial distribution is p(x) = nCxp x q n-x p(1) = 5C 1 pq 4 = 0.4096 (1) p(2) = 5C 2 p 2 q 3 = 0.2048 (2) Dividing (2) by (1) 10p 2 q 3 = 0.2048 5pq 4 0.4096 2p = 1 q 2 2p = 1 (1 - p) 2 4p = 1- p p = 1/5. Example 6: If X and Y are independent Binomial variables with parameters B 1 (5, ½) and B 2 (7, ½), find P(X + Y = 3) DIT 111 PROBABILITY AND QUEUEING THEORY 49 NOTES Anna University Chennai Solution: Since X and Y are independent binomial variables with parameters (5, ½) and (7, ½), (X + Y) is a binomial variable with parameters (12, ½) Therefore the probability distribution of the binomial variable X + Y is given by, P[X + Y = x] = 12 Cx (½) x (½) 12-x P[X + Y = 3] = 12 C 3 (½) 3 (½) 9 = 55/ 704 Example 7: Ten coins are tossed simultaneously, find the probability of getting i) at least seven heads ii) exactly seven heads iii) at most seven heads. Solution: n = 10 p = probability of head in the toss of a coin = ½ q = 1 –p = 1 – ½ = ½ p(x) = nCxp x q n-x p(x) = 10Cxp x q 10 – x p(x) = 10Cx(½) x (½) 10-x = 10Cx (1/2 10 ) i)Probability of getting at least 7 heads P(x¡7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = (½) 10 [10C 7 + 10C 8 + 10C 9 + 10C 10 ] = [120 + 45 +10 +1] = 11 1024 64 ii)Probability of getting exactly 7 heads = P(7) = 10C 7 (½) 10 = 15 128 iii)Probability of getting at the most 7 heads P(x , 7) = P(X = 0) + P(X = 1) + P(X = 2) + - - - - + P(X = 7) = 1 – [P(X = 8) + P(X = 9) + P(X = 10)] = 1 – . 1 . [10C 8 + 10C 9 + 10C 10 ] 2 10 = 1 – 56/1024 = 121/128. Example 8: If 10% of the screws produced by an automatic machine are defective, find the probability that out of 20 screws selected at random, there are DIT 111 PROBABILITY AND QUEUEING THEORY 50 NOTES Anna University Chennai i)exactly two defectives ii)at most three defectives iii)at least two defectives iv)between one and three defectives(inclusive). Find also mean and variance. Solution: p = probability that a screw is defective = 0.1 q = 0.9 p(x) = 20Cx(1/10) x (9/10) 20-x x = 0,1,2, - - - 20 i)probability exactly two are defectives p(x) = 20C 2 (1/10) 2 (9/10) 18 = 190 .9 18 10 20 ii) probability that there are at most three defectives = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = (9/10) 20 + 20C 1 (1/10)(9/10) 19 + 20C 2 (1/10) 2 (9/10) 18 + 20C 3 (1/10) 3 (9/10) 17 = (9/10) 17 [(9/10) 3 + 20(1/10)(9/10) 2 + 190 (1/10) 2 (9/10) + 1140(1/10) 3 ] = 9 17 /10 20 [729 + 1620 + 1710 + 1140] iii)Probability that there are atleast two defectives P(X ¡2) = 1 – P(X<2) = 1 – [P(X = 0) + P(X = 1)] = 1 – [ (9/10) 20 + 20C 1 (1/10)(9/10) 19 ] = 1 – (9 19 /10 20 )(9 +20) = 1 – (9 19 /10 20 )(29) iv)Probability that the number of defectives is between one and three (inclusive) P(1,X,3) = P(X = 1) + P(X = 2) + P(X = 3) = 20C 1 (1/10)(9/10) 19 + 20C 2 (1/10) 2 (9/10) 18 + 20C 3 (1/10) 3 (9/10) 17 = (9 17 /10 20 )(4470) v)Mean = np = 20 .(1/10) = 2 variance = npq = 20.(1/10)(9/10) = 9/5 Example 9:The probability of a successful rocket launching is p. If launching attempts are made until 3 successful launchings have occurred, what is the probability that exactly 5 attempts are necessary? what is the probability that fewer than 5attempts will be necessary? iii)If the launching attempts are made until 3 consecutive successful launchings occur, what are the probabilities? DIT 111 PROBABILITY AND QUEUEING THEORY 51 NOTES Anna University Chennai Solution: p(x) = nCxp x q n-x i)Exactly 5 attempts will be required to get 3 successes, if 2 successes occur in the first four attempts and third success in the fifth attempt. Therefore P(exactly 5 attempts are required ) = P(2 successes in 4 attempts) x P(success in the single fifth attempt) Now, P(2 successes in 4 attempts = 4C 2 p 2 q 2 (since it follows binomial distribution) P(success in the single fifth attempt) = p (given) Therefore P(exactly 5 attempts are required ) = 4C 2 p 2 q 2 x p = 6p 3 q 2 . ii)P(fewer than 5attempts are required) = P(exactly 3 or 4 attempts are required) = [P(2 successes in first 2 attempts) x P(success in the 3 rd attempt)] + [P(2 successes in first 3 attempts) x P(success in the 4 th attempt] = 2C 2 p 2 q 0 x p + 3C 2 p 2 q 1 x p = p 3 + 3p 3 q = p 3 (1 + 3q) iii)Five attempts will be required to get 3 consecutive successes, if the first 2 attempts result in failures and the last 3 attempts result in success. Therefore required probability = q.q.p.p.p = q 2 p 3 Three attempts will be required to get 3 consecutive success, if each attempt result in a success. Therefore required probability =p.p.p = p 3 Four attempts will be required to get 3 consecutive success, if first attempt result in a failure and the remaining attempts result in a success each. Therefore required probability = q.p.p.p = qp 3 Therefore, if the launching attempts are made until 3 consecutive successful launchings occur, probability = q 2 p 3 + p 3 + qp 3 = p 3 (1 + q) Example 10: Fit a binomial distribution for the following data: x: 0 1 2 3 4 5 6 total f: 5 18 28 12 7 6 4 80 Solution: To find the binomial distribution which fits the given data, we require N, n and p. We assume N = total frequency = 80 and from the given data n = 6. To find p : We know mean = np. From the given data we will find the mean and then equate to np so that we can find the value of p. DIT 111 PROBABILITY AND QUEUEING THEORY 52 NOTES Anna University Chennai x : 0 1 2 3 4 5 6 total f : 5 18 28 12 7 6 4 80 fx: 0 18 56 36 28 30 24 192 mean = Σ fx = 192 Σ f 80 = 2.4 i.e, np = 2.4, hence p = 0.4 and q = 0.6 Now we will find the theoretical frequencies which is given by Np(x) We have p(x) = nCxp x q n-x When x = 0, 80p(0) = 6C 0 p 0 q 6 = 3.73 When x = 1, 80p(0) = 6C 1 p 1 q 5 = 14.93 When x = 2, 80p(0) = 6C 2 p 2 q 4 = 24.88 When x = 3, 80p(0) = 6C 3 p 3 q 3 = 22.12 When x = 4, 80p(0) = 6C 4 p 4 q 2 = 11.06 When x = 5, 80p(0) = 6C 5 p 5 q 1 = 2.95 When x = 6, 80p(0) = 6C 6 p 6 q 0 = 0.33 Thus we have x: 0 1 2 3 4 5 6 f: 3.73 14.93 24.88 22.12 11.06 2.95 0.33 Converting these values into whole numbers consistent with the condition that the total frequency distribution is 80, the corresponding binomial frequency distribution is as follows x: 0 1 2 3 4 5 6 Theoretical f: 4 15 25 22 11 3 0 Example 11: Assume that half of the population is vegetarian so that the chance of an individual being a vegetarian is ½ . Assuming that 100 investigators take samples of 10 individuals each to see whether they are vegetarians, how many investigators would you expect the reports that 3 people or less were vegetarians. Solution p = probability that an individual is a vegetarian = ½ q = 1 – p = ½ n = Number of individuals for each investigator = 10 N = Number of investigators The expected number of investigators reporting x persons as vegetarians = N. n C x p x q n – x = 100. 10 C x ( ½ ) x . (½ ) 10 – x = 100. 10 C x (½ ) 10 DIT 111 PROBABILITY AND QUEUEING THEORY 53 NOTES Anna University Chennai Therefore the number of investigators reporting three or less as vegetarians = N [ p (0) + p(1) + p(2) + p(3) ] = 100. (½ ) 10 [ 10C 0 + 10C 1 + 10C 2 + 10C 3 ] = (100/1024) [ 1 + 10 + 45 +120 ] = 176 x 100 = 275 = 17 1024 16 Example 12: In a certain town, 20% samples of the population is literate and assume that 200 investigators take samples of ten individuals to see whether they are literate. How many investigators would you expect to report that three people or less are literates in the samples. Solution p = probability that an individual is literate = 20% = 0.2 q = 1 – p = 0.8 n = 10 p (x) = 10C x (0.2) x ( 0.8) 10 – x Therefore number of investigators reporting 3 or less as literate = N [ p(0) + p(1) + p(2) + p(3) ] = 200 [ (0.8)10 + 10C 1 (0.2) (0.8) 9 + 10C 2 (0.2) 2 (0.8) 8 + 10C 3 (0.2) 3 (0.8) 7 ] = 176 Example 13An irregular six faced die is thrown and the probability that in 10 throws it will give five even numbers is twice the probability that it will give four even numbers. How many times in 10,000 sets of 10 throws would you expect it to give no even numbers. Solution Let p = probability of getting an even number in a throw of a die. Therefore p(x) = 10C x p x q 10 - x Given p = (5) = 2 p (4) Therefore 10C 5 p 5 q 5 = 2.10 C 4 p 4 q 6 10C 5 . p = 2q 10C 4 DIT 111 PROBABILITY AND QUEUEING THEORY 54 NOTES Anna University Chennai 10 -5 + 1 . p = 2q . nC r = n – r + 1 5 nC r-1 r 6p = 10 ( 1 – p ) p = 5 , q = 3 8 8 Therefore p( x ) = 10C x (5/8) x (3/8) 10 - x , x = 0,1,2, . . .. 10 Number of times no even number occurs in 10,000 sets of 10 throws = 10,000 p(0) = 10,000 ( 3 /8 ) 10 = 1 1.7.2 POISSON DISTIBUTION The Poisson distribution is used to model the number of events occurring within a given time interval. The Poisson distribution is a discrete probability distribution that expresses the probability of a number of events occurring in a fixed period of time if these events occur with a known average rate, and are independent of the time since the last event. It is also called as counting random variable. 17.2.1 Poisson distribution If X is a discrete r v that can assume the values 0,1,2 - - -such that its probability mass function is given by P(X = x) = e -λ λ x ; x = 0,1,2 - - -: λ >0 x! then is said to follow a Poisson distribution with parameter λ or symbolically X is said to follow P(λ) 1.7.2.2 Properties: The number of outcomes occurring during a time interval is independent of the number that occurs in any other disjoint time interval. So it is memory less. The probability that a single outcome will occur during a very short time interval is proportional to the length of the time interval. It does not depend on the number of outcomes that occur outside this time interval. The probability that mere then one outcome will occur in such a short time interval is negligible. 1.7.2.3 Mean and variance of Poisson distribution E(X) = x r p r x DIT 111 PROBABILITY AND QUEUEING THEORY 55 NOTES Anna University Chennai · = x e -λ λ x (1) x = 0 x! · = λ e -λ λ x - 1 x = 0 (x - 1)! = λ e -λ e λ = λ (2) E(X 2 ) = x x 2 p x x · = {x(x - 1) + x} e -λ λ x x = 0 x! · = λ 2 e -λ λ x – 2 + λ x = 2 (x - 2)! = λ 2 e -λ e λ + λ = λ 2 + λ Var (X) = E(X 2 ) + { E(X)} 2 = λ 2 + λ - λ 2 = λ 1.7.2.4 MGF of Poisson distribution: · M X (t) = e tx e -λ λ x x = 0 x! 8 = e -λ (λe t ) x x = 0 x! = e -λ [ 1 + λe t + (λe t ) 2 + - - ] 1! 2! -λ  (λ e t ) = e e or e -λ exp(λe t ) = exp(λ(e t -1)) 1.7.2.5 Recurrence formula for Poisson distribution: P(X = x) = e -λ λ x ; x = 0,1,2 - - -: λ >0 x! P(X = x + 1) = e -λ λ x+1 (x +1)! DIT 111 PROBABILITY AND QUEUEING THEORY 56 NOTES Anna University Chennai P(X = x + 1) = λ . P(X = x) (x +1) P(X = x + 1) = λ P(X = x) , x>0 (x +1) 1.7.2.6 Additive property of Poisson Random Variables If X and Y are two independent Poisson Random Variables with means λ 1 & λ 2 Respectively, then X + Y is also a Poisson Random Variable with mean λ 1 + λ 2 . 1.7.2.7 Poisson distribution as a limiting form of Binomial Distribution Poisson distribution as a limiting form of Binomial Distribution under the following conditions. i) n, the number of trials is indefinitely large i.e, n ¦ · ii) p, the constant probability of success in each trial is very small, i.e, p  ·. iii) np =λ is finite or p = λ/n and q = 1 - λ/n, where λ is a positive real number. Proof: If X is a binomially distributed r v with parameters n and p then, P(X = x ) = nCx p x q n-x , x = 0,1,2,- - - = n(n-1) (n-2) - - - (n-(x -1)) p x (1 -p) n -x x! = n(n-1) (n-2) - - - (n-(x -1)) (λ/n) x (1 – (λ/n)) n –x (on putting p = λ/n) x! = λ x 1 (1 – 1/n)(1 – 2/n) - - - (1 – (x -1)/n) (1 - λ/n ) n . (1 - λ/n) -x x! lim(P(X=x) = λ x lim (1–1/n)(1–2/n)---(1– (x-1)/n).lim(1- λ/n ) n .lim(1- λ/n) -x n¦· x! n · n · n · np = λ = finite = λ x e -λ , which is the probability mass function of Poisson RV x! since lim (1 – k/n) = 1 when k is finite, lim(1 - λ/n ) n = e -λ , lim(1 - λ/n) -x =1 n · n · n · Therefore we may compute binomial probabilities approximately by using the corresponding Poisson probabilities, whenever n is large and p is small. DIT 111 PROBABILITY AND QUEUEING THEORY 57 NOTES Anna University Chennai Some examples of Poisson variables are i)the number trains arriving in a railway station in a given time interval. ii)the number of alpha particles emitted by a radio active source in a given time interval iii) the number of accidents reported in a town per day. iv)to count the number of causalities in insurance problems. Example1: The number of accidents in a year to taxi-drivers in a city follows a Poisson distribution with mean equal to 3. Out of 1000 taxi drivers, find approximately the number of drivers with i) no accidents in a year ii) more than 3 accidents in a year Solution: Here mean = λ = 3 The probability function is given by P(X = x) = e -λ λ x ; x = 0,1,2 - - -: λ >0 x! where is the number of accidents in a year. i)Probability of no accidents in a year P(0) = e -3 = 0.0498 therefore out of 1000 drivers, the number of drivers with no accidents in a year is 1000 x 0.0498= 49.8 [ 50. ii) Probability of more than 3 accidents in a year = 1 - Probability of not more than 3 accidents in a year = 1 – P(X ,3) = 1 – [P(0) + P(1) + P(2) + P(3) ] P(1) = e -3 .3 P(2) = e -3 .3 2 /2! = 4.5e -3 P(3) = e -3 .3 3 /2! = 4.5e -3 Therefore Probability of more than 3 accidents in a year = 1 – [P(0) + P(1) + P(2) + P(3) ] = 1 – e -3 [1 + 3 + 4.5 + 4.5] = 1 - 0.6474 = 0.3526. therefore out of 1000 drivers, the number of drivers with more than 3 accidents in a year is 1000 x 0.3526 = 352.6 [ 353. Example 2: Fit a Poisson distribution for the following data. x: 0 1 2 3 4 5 total f(x): 142 156 69 27 5 1 400 Solution To find the poisson distribution which fits the given data, we require N and λ. We assume N = total frequency = 400 To find λ : We know mean = λ DIT 111 PROBABILITY AND QUEUEING THEORY 58 NOTES Anna University Chennai . From the given data we will find the mean and hence λ. x: 0 1 2 3 4 5 f : 142 156 69 27 5 1 fx: : 0 156 138 81 20 5 mean = fx = 400 = 1 f 400 λ = 1 Theoretical frequencies are given by N e -λ λ x , x = 0, 1,2,3,4,5 where N = 400 x! Thus we get x: 0 1 2 3 4 5 f: 147.15 147.15 73.58 24.53 6.13 1.23 Converting these values into whole numbers consistent with the condition that the total frequency distribution is 400, the corresponding binomial frequency distribution is as follows x: 0 1 2 3 4 5 Theoretical f: 147 147 74 25 6 1 Example 3: It is known that the probability of an item produced by a certain machine will be effective is 0.05. If the produced items are sent to the market in packets of 20,find the number of packets containing at least , exactly and atmost 2 defective items in a consignment of 1000 packets using i) binomial distribution ii) poisson approximation to binomial distribution Solution: Using Binomial i) p = probability that an item is defective = 0.05 q = 0.95 n = Number of independent trials considered P(X = x) = nCxp x q n-x a) P (exactly 2 defectives) = P(X = 2) = 20C 2 p 2 q 18 = 0.1887 If N is the number of packets, each packet containing 20 items, then the number of packets containing exactly 2 defectives is given by N x P(X = 2) = 1000 x 0.1887 = 189, approximately b)P(at least 2 defectives) = P(X ¡ 2) = 1 – [p(0) + p(1)] = 1 –[20C 0 (0.05) 0 (0.95) 20 + 20C 1 (0.05) 1 (0.95) 19 ] = 0.2641 DIT 111 PROBABILITY AND QUEUEING THEORY 59 NOTES Anna University Chennai If N is the number of packets, each packet containing 20 items, then the number of packets containing at least 2 defectives is given by N x P(X ¡ 2) = 1000 x 0.2641 = 264, approximately c) P(at most 2 defectives) = P(X ,2) = p(0) + p(1) + p(2) = 20C 0 (0.05) 0 (0.95) 20 + 20C 1 (0.05) 1 (0.95) 19 + 20C 2 (0.05) 2 (0.95) 18 = 0.9246 If N is the number of packets, each packet containing 20 items, then the number of packets containing atmost 2 defectives is given by N x P(X ,2) = 1000 x 0.9246 = 925, approximately Using Poisson ii) since p = 0.05 is very small and n = 20 is sufficiently large, binomial distribution may be approximated by poisson distribution with parameter λ = np = 1 therefore P(X = x) = e -λ λ x = e -1 x! 1! a) P (exactly 2 defectives) = P(X ¡ 2) = e -1 = 0.1839 2! If N is the number of packets, each packet containing 20 items, then the number of packets containing exactly 2 defectives is given by N x P(X = 2) = 1000 x 0.1839 = 184, approximately b) P(at least 2 defectives) = P(X ¡2) = 1 – [p(0) + p(1)] = 1 – [e -1 + e -1 ] = 0.2642 If N is the number of packets, each packet containing 20 items, then the number of packets containing at least 2 defectives is given by N x P(X ¡ 2) = 1000 x 0.2642 = 264, approximately c) P(at most 2 defectives) = P(X,2) = p(0) + p(1) + p(2) = 0.9197 If N is the number of packets, each packet containing 20 items, then the number of packets containing atmost 2 defectives is given by N x P(X,2) = 1000 x 0.9197 = 920, approximately Example 4: Assume that the number of planes crossing Indian border during war between 5p.m and 6p.m is a Poisson random variable with parameter 3 and the number between 6p.m and 7p.m is a Poisson random variable with parameter 4. If these two DIT 111 PROBABILITY AND QUEUEING THEORY 60 NOTES Anna University Chennai random variables are independent what is the probability that more than 5 planes cross the border between 5p.m and 7 p.m? Solution: Let X 1 be the number of planes crossing the border between 5p.m and 6p.m X 2 be the number of planes crossing the border between 6p.m and 7p.m. Since X 1 and X 2 are independent Poisson random variables with parameters 3 and 4 respectively, X 1 + X 2 is a Poisson random variable with parameter 7. P(X 1 + X 2 > 5) = 1 – P(X 1 + X 2 ,5) 5 = 1- Σ e -7 7 x x = 0 x! = 1 – 0.3007 = 0.6993 Example 5: After correcting the proofs of the first 50 pages of a book, it is found that on the average there are 3 errors per 5 pages. Use Poisson probabilities and estimate the number of pages with 0,1,2,3 errors in the whole book of 1000 pages(e -6 = 0.5488) Solution: λ = mean = average no : of errors per page = 3/5 = 0.6 The probability that there are x errors per page is p(x) = e -λ λ x = e -0.6 (0.6) x , x = 0,1,2 – - - — - x! x! i)Number of pages containing no error N x p(0) = 1000 e -0.6 = 1000 x .5488 ˜ 549 pages ii)Number of pages containing one error N x p(1) = 1000 e -0.6 x 0.6 = 329 pages 1! iii)Number of pages containing 2 error N x p(1) = 1000 e -0.6 x (0.6) 2 = 1000 x 98.7 ˜ 99 pages 2! iv)Number of pages containing 3 error N x p(1) = 1000 e -0.6 x (0.6) 3 = 20 pages 3! Example 6: Assume that the chance of an individual coal miner being killed in a Smine accident during a year is 1/1400 . Use Poisson distribution to calculate the probability that in a mine employing 350 miners, there will be atleast one fatal accident in a year (e -.25 = 0.7788) Solution: p(x) = e -λ λ x , x = 0,1,2 – - - — - x! DIT 111 PROBABILITY AND QUEUEING THEORY 61 NOTES Anna University Chennai p = 1/1400; n = 350 λ = np = 350/1400 = 0.25 p(x) = e -0.25 (0.25) x , x = 0,1,2 – - - — - x! The probability that there will be atleast one fatal accident = P(X=1) = 1 – P(0) = 1- e -0.25 = 1- 0.7788= 0.2212 Example 7: If X and Y are independent poisson random variables, show hat the conditional distribution of X, given the value of X + Y, is a Binomial distribution. Solution: Let X and Y follow Poisson distributions with parameters λ 1 & λ 2 respectively. Now P[X =x / (X + Y) = n] = P[X = x and(X + Y) = n] = P[X = x ;Y = n –x] P[ (X + Y) = n] P[ (X + Y) = n] = P[X = x ].P[X = n – x] (by independence of X and Y) P[ (X + Y) = n] = [e - λ1 . λ 1 x / x!][ e - λ2 . λ 2 n-x /(n - x)!] e -( λ1+ λ2) .( λ 1 + λ 2 ) n / n! = n! λ 1 r λ 2 n-r x!(n – x)! (λ 1 + λ 2 ) (λ 1 + λ 2 ) = nC r p r q n-r Where p = λ 1 r λ 2 n-r (λ 1 + λ 2 ) q = (λ 1 + λ 2 ) Example 8: One-fifth of the blades produced by a blade manufacturing factory turn out to be defective. The blades are supplied in packets of 10. Use Poisson distribution to calculate the approximate number of packets containing no defective and two defective blades in a consignment of 1,00,000 packets. Solution: p = (1/5)/100 = .002 n = 10 λ = np = 0.2 p(x) = e -λ λ x = e -0.2 (0.2) x x! x! i)Number of packets containing no defective DIT 111 PROBABILITY AND QUEUEING THEORY 62 NOTES Anna University Chennai = Np(0) = 1,00,000e- 0.2 = 98020 ii) Number of packets containing one defective = Np(1) = 1,00,000e- 0.2 (0.2) = 1960 1! iii) Number of packets containing two defectives = Np(2) = 1,00,000e- 0.2 (0.2) 2 = 20 2! 1.7.3 GEOMETRIC DISTRIBUTION Suppose that I am at a party and I start asking girls to dance. Let X be the number of girls that I need to ask in order to find a partner. If the first girl accepts, then X=1. If the first girl declines but the next girl accepts, then X=2. And so on. When X=n, it means that I failed on the first n-1 tries and succeeded on the nth try. My probability of failing on the first try is (1-p). My probabilty of failing on the first two tries is (1-p)(1-p). My probability of failing on the first n-1 tries is (1-p) n-1 . Then, my probability of succeeding on the nth try is p. Thus, we have P(X = n) = (1 - p) n-1 p This is known as the geometric distribution 17.3.1 Geometric distribution Definition : Let RV X denote the number of trial of a random experiment required to obtain the first success ( occurrence of an event A ). Obliviously X can assume the values 1,2,3 …. Now X= x, if and only if the first ( x-1) trails result in failure ( occurrence of Ä ) and the x th trial results in success ( occurrence of A ) Hence P ( X = x ) = q x - 1 p; x = 1,2,3…., 8 Where P ( A ) = p and P (Ā) = q If X is a discrete RV that can assume the values 1,2,3,….. 8such that its probability mass function is given by P ( X = x ) = q x - 1 p; x = 1,2,3…., 8where p + q = 1 DIT 111 PROBABILITY AND QUEUEING THEORY 63 NOTES Anna University Chennai Then X is said to follow a geometric distribution. Note – Geometrics distribution is legitimate probability distribution since · · Σ P (X = x ) = Σ q x – 1 p x =1 x =1 = p ( 1 + q + q 2 + …. ·) p = 1 1 - q 1.7.3.2 Mean and variance of Geometric distribution E ( X ) = Σ x x P x x · = Σ xq x – 1 p x = 1 = p ( 1 + 2q + 3q 2 + ..... + ·) = p ( 1 – q ) -2 = 1/p E ( X 2 ) = Σ x 2 p x x · = Σ x 2 q x – 1 p x = 1 · = p Σ { x ( x + 1 ) – x} q x - 1 x=1 = p [{1 X 2 + 3 X 3 q + 3 X 4 q 2 + ..... + ·} – { 1 + 2q + 3q 2 + ...... + ·}] = p [ 2 ( 1 – q ) -3 - ( 1 – q ) -2 ] = p { 2/p 3 - 1/p 2 } = 1/p 2 ( 2 – p ) = 1/p 2 ( 1 + q ) Var ( X ) = E ( X 2 ) - { E (X) } 2 DIT 111 PROBABILITY AND QUEUEING THEORY 64 NOTES Anna University Chennai = 1/p 2 ( 1 + q ) - 1/p 2 = q/p 2 Note – Sometimes the probability mass function of a geometric RV X is taken as P ( X = x ) = q x p; x = 0,1,2,……. · where p + q = 1 If this definition is assumed then E ( X ) = q/p and Var ( X ) = q/p 2 1.7.3.3 MGF of Geometric distribution: · Mx(t) = E(e tx ) = Σ e tx p q x-1 x = 1 = pe t Σ(e t q) x-1 = pe t (1 - qe t ) -1 = pe t 1 - qe t 1.7.3.4 Recurrence formula for Geometric distribution We have P ( X = x ) = q x - 1 p P ( X = x + 1 ) = q x p So P ( X = x + 1) = q P ( X = x ) P ( X = x + 1) = q P ( X = x ) Example:1 If the probability is 0.05 that a certain kind of measuring device will show excessive drift, what is the probability that the sixth of these measuring devices tested will be the first to show excessive drift? Solution: If the sixth device should show excessive drift then there should be 5 failures before the sixth trial, which is geometric distribution. P ( X = x ) = q x - 1 p; x = 1,2,3…., · P ( X = 6) = q 6 - 1 p Given p = 0.05, q = 0.95 P ( X = 6) =(0.95) 5 0.05 = 0.039 DIT 111 PROBABILITY AND QUEUEING THEORY 65 NOTES Anna University Chennai Example:2 A and B shoot independently until each has hit its own target. The probabilities of their hitting the target at each shot are 3/5 and 5/7 respectively. Find the probability that B will require more shots than A. Solution: Let X denote the number of trials required by A to get his first success. p = 3/5 q = 2/5 Then X follows a geometric distribution given by P ( X = x ) = q x - 1 p; x = 1,2,3…., · = 3/5. (2/5) x – 1 ; x = 1,2,3…., · Let Y denote the number of trials required by B to get his first success. p = 5/7 q = 2/7 Then X follows a geometric distribution given by P ( Y = x ) = q x - 1 p; x = 1,2,3…., · = 5/7. (2/7) x – 1 ; x = 1,2,3…., · Probability that B will require more shots than A. i.e, B requires more trials to get his first success than A requires to get his first success . i.e, probability of A getting his success in the x th trial and B getting his success in x + 1 th or x + 2 th or - - - trial. · = Σ P[X = x and Y = x + 1 th or x + 2 th or - - - · ] x = 1 · = Σ P[X = x ].P[Y = x + 1 th or x + 2 th or - - - · ] x = 1 · · = Σ (3/5)(2/5) x-1 . Σ (5/7)(2/7) x + k -1 x = 1 k = 1 · · = (3/7) Σ (2/5) x-1 Σ (2/7) k (2/7) x-1 x = 1 k = 1 · · = (3/7) Σ (2/5) x-1 (2/7) x-1 . Σ (2/7) k x = 1 k = 1 · · = (3/7) Σ (4/35) x-1 Σ (2/7) k x = 1 k = 1 · = (3/7) Σ (4/35) x-1 (2/7) x = 1 1 – 2/7 · = 6/35 Σ (4/35) x-1 = (6/35) 1 x = 1 1 - 4/35 = 6/31 DIT 111 PROBABILITY AND QUEUEING THEORY 66 NOTES Anna University Chennai Example:3 A die is thrown until 1 appears. Assuming that the throws are independent and the probability of getting 1 is p, find the value of p so that the probability that an odd number of throws is required is equal to 0.6. Can you find a value of p so that the probability is 0.5 that an odd number of tosses is required? Solution: Let X denote the number of throws required to get the first success (getting 1). Then the distribution of X is geometric P ( X = x ) = q x - 1 p; x = 1,2,3…., · P(X = an odd number) = P( X = 1 or 3 or 5 - - - -) · = Σ P(X = 2x - 1) x = 1 · = Σ pq 2x-2 x = 1 · = (p/q 2 ) Σ q 2k x = 1 = (p/q 2 )[q 2 + q 4 + - - ] = (p/q 2 )(q 2 /1-q 2 ) = p/(1 +q) Given 1/1 + q = 0.6 That is1/(2-p) = 0.6 => p = 1/3 If 1/1 + q = 0.5 then 1/(2-p) = 0.5 => p = 0 P = 0 is meaningless because · P(X = an odd number) = Σ pq 2x-2 = 0 x = 1 Hence the value of p cannot be found. Example 4: Establish memoryless property of geometric distributions, that is , if X is a discrete random variable following a Geometric distribution, then P{X>m +n / X>m} = P{X>n}, where m and n are any two positive integers. Prove the converse also, if it is true. Since X follows geometric distribution, P ( X = x ) = q x - 1 p; x = 1,2,3…., ·, p + q = 1 · P(X>k) = Σ q x - 1 p = p(q k + q k+1 + q k+2 + - - - + ·) x = k + 1 = pq k = q k. 1 –q P{X>m +n / X>m} = P{X>m +n and X>m} P{X>m} DIT 111 PROBABILITY AND QUEUEING THEORY 67 NOTES Anna University Chennai = P{X>m +n} = q m + n = q n = P(X>n) P{X>m} q m The converse of the above result is also true i.e, if P{X > m+n / X>m} = P{X>n}, where m and n are an y two positive integers, then X follows a geometric distribution. Since X takes the values 1,2,3, - - - -, P{X =1} = 1 Let P(>1) = q Now P{X = (x +1)} = P(X > x) - P(X > (x +1)) (1) P{X = (x +1)} = 1 – P(X > (x +1)) P(X > x) P(X > x) = 1 – P{X > (x+1)/X>x} = 1 – P{X > 1} = 1 – q P{X = (x +1)} = (1 - q) P(X > x) (2) = (1 - q)[P{ X > (x -1} - P{X = x} (from (1), on changing x to x -1 ) = (1 - q)[P{ X > (x -1} - (1 - q) P(X > x-1)] (from (2), on changing x to x -1 ) = (1 - q)q P(X > x-1) = (1 - q)q 2 P(X > x-2) = (1 - q)q x-1 P(X > 1) P{X = (x +1)} = (1 - q)q x P{X = x } = (1 - q)q x-1 where p = 1 – q and x = 1,2 -, - That is X follows geometric distribution. Have you understood ? Say true or false. Justify your answer. 1.Binomial distribution id continuous . 2.For binomial distribution variance < mean. 3.Mean and variance are different for Poisson distribution. 4.Poisson distribution is a symmetrical distribution. 5.Mean is always greater or equal to the variance for a geometric distribution. 6. Geometric distribution has no memory. (Answers: 1.False, 2.True, 3.False. 4.False,5.False, 6.True) DIT 111 PROBABILITY AND QUEUEING THEORY 68 NOTES Anna University Chennai Short answer questions 1. Derive the moment generating functions of all the distributions discussed above. 2. Derive the mean and variance of Poisson distribution. 3. State the additive property of Poisson distribution Try yourself ! 1. Determine the binomial distribution for which the mean is 4 and variance is 3. (Solution: 16 Cx (1/4) x (3/4) 16 – x , x = 0,1,2,- - -) 2.A and B playas game in which their chance of winning is in the ratio 3:2. Find A’s chance of winning atleast three games out of five games played. (Solution: 0.68) 3.If X is a poisson variate such that P(x =2) = 9 P(X = 4) + 90 P( X = 6), find the Variance. (Solution: 1 ) 4.A manufacturer of cotter pins knows that 5% of the product is defective. If he sells cotter pins the boxes of 100 and guarantees that not more than 4 pins will be defective. What is the approximate probability that a box will fail to meet the guaranteed quality. (Solution: P(X>4) = 0.5620) 5.If X is a geometric variate taking values 1,2, - - · find P(X is odd) (Solution: 1/(1 +q)) 6.If the probability that an applicant for a driver’s license will pass the road test on any given trial is 0.8, what is the probability that he will finally pass the test a) on the fourth trial b) in fewer than 4 trials? (Solution: 0.0064, 0.9984) 1.8 CONTINUOUS DISTRIBUTION 1.8.1 UNIFORM OR RECTANGULAR DISTRBUTION A uniform distribution is a distribution of a continuous variable in which the probability of X falling within a given interval is proportional to the size of the interval. For example, if X is uniformly distributed between 0 and 1, then the probability that X will be between 0.3 and 0.4 is .1, because there are ten intervals of width .1 each. The probability of X falling between 0.1 and 0.25 is .15. If X is uniformly distributed between 0 and 2, then what is the probability that X will fall between 0.3 and 0.4? Between 0.1 and 0.25? DIT 111 PROBABILITY AND QUEUEING THEORY 69 NOTES Anna University Chennai Uniform distributions do not occur very often in nature. However, random number generators often are built to simulate the uniform distribution. We can use a uniform random number generator to determine the winner of a raffle. Suppose that we have 247 entries, numbered one through 247. We can choose a random number between 0 and 1, multiply it by 247, and then round it to the the nearest integer in order to pick the winner. 1.8.1.1 Uniform distribution A continuous RV X is said to follow a uniform rectangular distribution over an interval (a,b) if its pdf is given by f(x) = 1/(b - a) a < x < b 0, otherwise Here a and b (b>a) are the parameters of the distributions . Distribution function x x F(x) = If(x)dx = I 1/(b - a) dx = (x - a)/(b-a) -· a Hence F(x) = 0 ; x < a (x - a)/(b-a); 0 < x < b 1 ; x > b 1.8.1.2 Mean and Variance · b Mean = E(X) == Ixf(x)dx = I [x/(b - a)]dx -· a b = 1/ (b-a) [x 2 /2] = b 2 – a 2 = b + a a 2(b - a) 2 · b E(X 2 ) == I x 2 f(x)dx = I [x 2 /(b - a)]dx -· a b = 1/ (b-a) [x 3 /2] = b 3 – a 3 = b 2 + ab + a 2 a 3(b - a) 3 Varaiance = E(X 2 ) – (E(X)) 2 = b 2 + ab + a 2 – (a 2 + 2ab + b 2 ) = b 2 – 2ab + a 2 3 4 12 = (b - a) 2 12 DIT 111 PROBABILITY AND QUEUEING THEORY 70 NOTES Anna University Chennai Mean = (b +a)/2; Variance = [(b - a) 2 ]/12 1.8.1.3 Moment Generating function b b Mx(t) = E(e tx ) = 1 I e tx dx = 1 e tx = e bt - e at (b – a) a (b – a) t b - a a Moments b b µ r ’ = E(X r ) = 1 I x r dx = 1 x r+1 = b r+1 – a r+1 (b – a) a (b – a) r +1 (r +1)(b – a) A Example 1: If X is uniformly distributed over (-3,3), find the probability that a)X<2 b)|x|<2 , c)|X - 2|<2 and find k such that P[X>k] = 1/3. Solution: f(x) = 1/(b - a) a < x < b 0, otherwise f(x) = 1/6 -3 < x < 3 0, otherwise 2 2 a) P(X<2) = I 1/6 dx = x/6 = 5/6 -3 -3 2 2 b) P(|X|<2) = P(-2 < x <2 ) = I 1/6 dx = x/6 = 4/6 = 2/3 -2 -2 c) P(|X-2|<2) = P(-2 < x-2 <2) = P(0< x< 4) = P(0 <x< 3) [since interval is (-3,3)] 3 3 I 1/6 dx = x/6 = 3/6 = ½ 0 0 d) P(X>k) = 1/3 3 3 => I 1/6 dx = x/6 = (3 – k)/6 = 1/3 (given) k k => 3 – k = 2 => k = 1 Example 2: A passenger arrives at a bus stop at 10am knowing that the bus will arrive at some time uniformly distributed between 10am and 10.30am. What is the probability that he will have to wait longer than 10min? If at 10.15am the bus has not arrived, what is the probability that he will have to wait atleast 10 additional minutes? DIT 111 PROBABILITY AND QUEUEING THEORY 71 NOTES Anna University Chennai Solution: Let X denote the waiting time. Then pdf of X is f(x) = 1 0 < x < 30 30 = 0 otherwise P ( he will have to wait longer than 10 min) = P( X > 10 ) 30 = I dx = 2 10 30 3 P ( he has to wait 25 min / he has already waited 15 min ) = P ( X > 25 / X > 15 ) = P ( X > 251X > 15 ) P (X > 15 ) = P ( X > 25 ) P ( X > 15 ) 30 = I dx 25 30 30 I dx 15 30 = 5 = 1 15 3 Example 3: Show that the mgf about origin for the rectangular distribution on (-a,a) is 1 sinh at. Also show that moments of even order are given by µ 2n = a 2n and all (2n + 1) moments of odd order vanish ( i.e µ 2n +1 = 0 ) Solution: Mgf about origin is given by DIT 111 PROBABILITY AND QUEUEING THEORY 72 NOTES Anna University Chennai a M X (t) = E [ e tX ] = 1 I e tx dx 2a -a = 1 ( e at – e –at ) 2at = sin hat at = 1 [at + (at ) 3 + (at ) 5 + ….. ] at 3! 5! = 1 + ( at ) 2 + (at ) 4 ….. 3! 5! Since there are no terms with odd powers of t in M X (t), all moments of odd order vanish i.e µ 2n +1 = 0. in particular µ 1 ' = 0 . Thus µ r = µ r 1 . Hence µ 2n +1 = 0 . The moments of even order are given by µ 2n +1 = coefficient of t 2n = a 2n . (2n)! 2n + 1 x Example 4: If RV has the density function f (x ) prove that y = f (x) = I f (x) dx has a -· rectangular distribution over ( 0 , 1 ). If f (x) = x-1 , 1 , x , 3 2 0 , otherwise Determine what interval for Y will correspond to the interval 1.1 = X = 2.9 Solution: The RV Y is defined as Y = F X (x) the distribution function of Y is F y (y ) = P ( Y , y ) = P ( F X (x) , y ) = P ( X , f -1 x (y) = F x [ F -1 x (y ) ] ( since P ( X , x ) = F x ( x) ) = y DIT 111 PROBABILITY AND QUEUEING THEORY 73 NOTES Anna University Chennai Thus the density function of y is given by F y (y) = d [ F Y (y) ] = 1 dy The range of Y is 0 , y , 1 since the range of Fx ( x) is ( 0, 1 ). Thus Y follows a uniform distribution in ( 0, 1 ) The distribution function of X is x F x ( x ) = I x – 1 dx = (x – 1 ) 2 1 2 4 Since Y = F x (x) , Y = 1 ( X – 1 ) 2 4 Thus when 1.1 , X , 2.9, 1 ( 1.1 – 1 ) 2 , Y , 1 ( 2.9 – 1 ) 2 4 4 0.0025 , Y , 0.9025. Example 5: Buses arrive at a specified stop at 15 min. intervals starting at 7 a m. i.e. they arrive at 7, 7:15. 7:30,7:45and so on . If a passenger arrives at the stop between 7 and 7.30 am, find the probability that he waits a)less than 5 min for a bus and b)atleast 12 min for a bus. Solution Let X denotes the time in minutes past 7 a.m, when the passenger arrives at the stop. Then X is uniformly distributed over (0,30) f(x) = 1 0 < x < 30 30 = 0 otherwise a) The passenger will have to wait less than 5 min. if he arrives at the stop between 7:10 and 7:15 or 7:15 and 7:30. Therefore the required probability= P(10<x<15) + P(25<x<30) 15 30 = I dx + I dx = 1/3 10 30 25 30 b)The passenger will have to wait atleast 12 minutes. if he arrives between 7 and 7:03 or 7:15 and 7:18 DIT 111 PROBABILITY AND QUEUEING THEORY 74 NOTES Anna University Chennai Therefore the required probability= P(0<x< 3) + P(15<x<18) 3 18 = I dx + I dx = 1/5 0 30 15 30 Example 6: Let two independent random variables X and Y have the geometric distribution. Show that the conditional distribution of X/(X + Y= k) is uniform. Solution: P(X = x) = P(Y = y) = pq x-1 P{X/(X + Y= k)} = P{X = x and (X + Y= k)} = P{X = x },P (Y= k - x)} = pq x-1 . pq k – x-1 k – 1 k – 1 Σ P{X = x },P (Y= k - x)} Σ pq x-1 . pq k –x-1 x = 1 x = 1 = q k – 2 = 1/(k – 1); x = 1,2,3, - - (k - 1) k - 1 Σ q k - 2 x = 1 Thus the conditional distribution of X , given that X + y = k , is a discrete uniform distribution. 1.8.2 EXPONENTIAL DISTRIBUTION 1.8.2.1Exponential distribution A continuous random variable X assuming non-negative values is said to have an exponential distribution with parameter λ> 0, if its pdf is given by f(x) = λe – λx x=0 0, otherwise Distribution function: The distribution function F(x) is given by x F(x) = I λe – λx dx = 1 - e – λx 0 Therefore F(x) = 1 - e – λx x ¡ 0 0 otherwise DIT 111 PROBABILITY AND QUEUEING THEORY 75 NOTES Anna University Chennai 1.8.2.2 Mean and variance · · µ r ’ = E[x r ] = I x r λe – λx dx = λ I x r e – λx dx 0 0 Put λx = y · µ r ’ = E[x r ] = λ I (y/ λ) r e – λx (dy/ λ) 0 = (1/ λ r ) \(r + 1) = r! / λ r Mean = E(X) = µ 1 ' = 1/ λ µ 2 ' = E(X 2 ) = 2 / λ 2 Var(X) = µ 2 = µ 2 ' – (µ 1 ') 2 = 2/ λ 2 – 1/ λ 2 = 1/ λ 2 Mean = 1/ λ and variance = 1/ λ 2 1.8.2.3 Moment generating function · Mx(t) = E[e tx ] = I e tx λe – λx dx 0 · –1 = λ e –(λ - t)x = λ = 1 – t -(λ - t) λ - t λ 0 1.8.2.4 Memoryless property If X is exponentially distributed then P(X>s +t/X>s) = P(X>t) for any s,t >0 · · Proof: P(X >s) = I λe – λx dx = –e – λx = e – λs 0 S P(X>s +t/X>s) = P(X>s +t 1 X>s) = e – λ(s + t) = e – λt = P(X>t) P(X > s) e – λs Example 1:If X is exponentially distributed prove that the probability that X exceeds its expected value is less than ½. Solution: Let X be exponentially distributed with parameter λ. Then f(x) = λe – λx x=0 E(X) = 1/λ DIT 111 PROBABILITY AND QUEUEING THEORY 76 NOTES Anna University Chennai · P(X > 1/λ) = I λe – λx dx = e –1 = 0.3679 < = ½ 1/λ Example 2: The time in hours required to repair a machine is exponentially distributed with parameter λ = ½ i)What is the probability that the repair time exceeds 2 hours. ii) what is the conditional probability that a repair takes atleast 10 hours given that its duration exceeds 9 hours. Solution: Given λ = ½ Let X denote the time to repair the machine. The density function of X is given by f(x) = λe – λx x=0 = ½ e –x/2 i) the probability that the repair time exceeds 2 hours. · · P(X>2) = I λe – λx = I ½ e –x/2 dx = e –1 = 0.3679 2 2 ii) the conditional probability that a repair takes atleast 10 hours given that its duration exceeds 9 hours is given by P(X>10/X>9) = P(X>9 + 1>X>9)= P(X >1) (using memoryless property) · = I ½ e –x/2 dx = e –0.5 = 0.6065 1 Example 3:The daily consumption of milk in excess of 20,000 liters in a town is approximate exponentially distributed with parameter 1/3000. The town has a daily stocks of 35,000 liters. What is the probability that of 2 days selected at random the stock is insufficient for both days? Solution: If Y denotes daily consumption of milk then X = Y – 20000 follows an exponential distribution with parameter 1/3000. Then f(x) = 1/3000 . e –x/3000 , x>0 P(stock insufficient for one day) = P(Y>35000) = P(X + 20000 > 35000) = P(X>15000) · = 1/3000 I e –x/3000 dx 15000 = e –5 P(Stock insufficient for 2 days) = (e –5 ) 2 e –10 DIT 111 PROBABILITY AND QUEUEING THEORY 77 NOTES Anna University Chennai Example 4: The mileage which car owners get with a certain kind of radial tire is a random variable having an exponential distribution with mean 40,000 km. Find the probabilities that one of these tires will last i) at least 20,000 km and ii) at most 30,000 km. Solution: Let X denotes the mileage obtained with the tire. Mean = 1/λ = 40,000 = > λ = 1/40000 The density function of X is given by f(x) = λe – λx x ¡ 0 = 1/40000 e –(1/40000)x i)The probability that one of these tires will last atleast 20,000 km is given by P(X=20,000) · = I 1/40000 e –(1/40000)x dx 20,000 · = – e –(1/40000)x = e –0.5 40000 20,000 ii)The probability that one of those tires will last atmost 30,000 km is given by P(X=30,000) 30,000 = I 1/40000 e –(1/40000)x dx 0 30,000 = – e –(1/40000)x = – e –0.75 +1 = 0.5270 40000 0 Example 5: A company decides to manufacture a specialized type of fuse whose lifetime is a exponential distribution. On a survey it was found that there were two processes by which the fuse may be manufactured. If a process I is used, the fuse made will have an expected length of 100hrs whereas those made by process II have an expected life length of 150 hrs. Process II is twice as costly per fuse process I which will cost Rs. 6 per fuse. Further more, if a fuse lasts for less than 200 hrs for which period it is guaranteed, a loss of Rs. 40 is assessed against the manufacturer. Which process should the company adopt? Solution: The density function of X is given by f(x) = λe – λx x ¡ 0 DIT 111 PROBABILITY AND QUEUEING THEORY 78 NOTES Anna University Chennai If process I is used , given Expected value as 100, i.e. 1/λ = 100 = >λ = 1/100 Then f(x) = (1/100)e – (x /100) x¡0 P(X¡200) · = I (1/100)e – (x /100) dx = e –2 200 P(X<200) = 1 - P(X¡200) = 1 -e –2 If process II is used , given Expected value as 150, i.e. 1/λ = 150 = >λ = 1/150 Then f(x) = (1/100)e – (x /100) x¡0 P(X¡200) · = I (1/150)e – (x /150) dx = e –4/3 200 P(X<200) = 1 - P(X¡200) = 1 -e –4/3 Let C 1 and C 2 be the costs per fuse corresponding to process I and II respectively. Then C 1 = 6 X¡200 46 X<200 (loss of Rs.40 + manufacturing cost Rs. 6) Therefore E(C 1 ) = 6P(X¡200) + 46 P(X < 200) = 6 e –2 + 46(1 – e —2 ) = 40.5866 Similarly, C 2 = 12 X¡200 52 X<200 (loss of Rs.40 + manufacturing cost Rs. 6) Therefore E(C 2 ) = 6P(X¡200) + 46 P(X < 200) = 12 e –4/3 + 52(1 – e —4/3 ) = 41.456 Since E(C 1 ) < E(C 2 ) process I should be adopted. 1.8.3 Normal or Gaussian distribution The normal distribution is almost the opposite of the uniform distribution. The uniform distribution is mathematically simple but occurs rarely in nature. The normal distribution is mathematically complex but occurs frequently in nature. 1.8.3.1 Normal distribution A continuous random variable X is said to follow a normal distribution or Gaussian distribution with parameters μ and σ, if its probability density function is given by DIT 111 PROBABILITY AND QUEUEING THEORY 79 NOTES Anna University Chennai f(x) = ; -· < x < · -· < µ < ·, σ>0 (1) Or f(x) = 1 exp(-(x - µ ) 2 /2σ 2 ) ; -· < x < · σ \2π -· < µ < ·, σ>0 Symbolically X follows N(μ, σ). Some times it is also given as N(μ, σ 2 ). 1.8.3.2 Standard Normal distribution The normal distribution N(0, 1) is called the standardized or simply the standard normal distribution, whose density function is given by Φ(z) = 1 exp(-z 2 /2) ; -· < z < · \2π This is obtained by putting μ = 0 and σ = 1 and by changing x and f respectively into z and Φ. If X has distribution N(μ, σ) and if Z = X - μ , then we can prove that Z has distribution N(0,1) σ 1.8.3.3 Normal Probability curve Normal probability curve 1.8.3.4 Characteristics of the Normal Distribution: 1. It is bell shaped and is symmetrical about its mean. 2. It is asymptotic to the axis, i.e., it extends indefinitely in either direction from the mean. 3. It is a continuous distribution. 4. It is a family of curves, i.e., every unique pair of mean and standard deviation defines a different normal distribution. Thus, the normal distribution is completely described by two parameters: mean and standard deviation. DIT 111 PROBABILITY AND QUEUEING THEORY 80 NOTES Anna University Chennai 5. Total area under the curve sums to 1, i.e., the area of the distribution on each side of the mean is 0.5. 6. It is unimodal, i.e., values mound up only in the center of the curve. 7. The probability that a random variable will have a value between any two points is equal to the area under the curve between those points. 1.8.3.5 Mean and variance of Normal distribution · E(X) = I x f(x) dx -· · = 1 I x exp(- (x - µ) 2 / 2 σ 2 ) dx σ\2π -· · = 1 I (µ + \2 σt) exp (-t 2 )dt (on putting t = x - µ ) \π -· σ\2 · · = µ I exp (-t 2 )dt + \2 σ I t exp (-t 2 )dt \π -· \π -· (the integrand in the second part is an odd function hence it reduces to 0) = µ \π = µ \π Mean = µ · E(X 2 ) = I x 2 f(x) dx -· · = 1 I x 2 exp(- (x - µ) 2 / 2 σ 2 ) dx σ\2π -· · = 1 I (µ + \2 σt) 2 exp (-t 2 )dt (on putting t = x - µ ) \π -· σ\2 · · · = 1 I µ 2 exp (-t 2 )dt + 2\2 µ σ I t exp (-t 2 )dt + 2 σ 2 I t 2 exp (-t 2 )dt \π -· -· -· · = µ 2 \π + 0 + 2 σ 2 2 I t.t exp (-t 2 )dt ; (since t 2 exp(-t 2 ) is an even function) \π 0 · = µ 2 + 2 σ 2 I t exp (-t 2 ) 2t dt \π 0 DIT 111 PROBABILITY AND QUEUEING THEORY 81 NOTES Anna University Chennai Put u = t 2 ; du = 2tdt · = µ 2 + 2 σ 2 I u 1/2 exp (-u) du \π 0 · = µ 2 + 2 σ 2 \(3/2) (by definition of gamma function \ n = Ix n-1 e -x dx) \π 0 = µ 2 + 2 σ 2 (1/2) | (1/2) ( | n = (n - 1) | (n -1)) \π = µ 2 + σ 2 \π ( | (1/2) = \π ) vπ = µ 2 + σ 2 Var (X) = E(X 2 ) –[ E(X)] 2 = σ 2 1 .8.3.6 Median of the normal distribution N (μ, σ) IF X is a continuous random variable with density function f(x), then M is called the median value of X, provided that M · I f(x) dx = I f(x) dx = ½ · M For the normal distribution N (μ, σ), the median is given by · I 1 exp(- (x - µ) 2 / 2 σ 2 ) dx = ½ M σ\2π µ · I 1 exp(- (x - µ) 2 / 2 σ 2 ) dx + I 1 exp(- (x - µ) 2 / 2 σ 2 ) dx = ½ M σ\2π μ σ\2π · But I 1 exp(- (x - µ) 2 / 2 σ 2 ) dx = 1 and the normal curve is symmetrical -· σ\2π · about x = µ I 1 exp(- (x - µ) 2 / 2 σ 2 ) dx = ½ μ σ\2π µ I 1 exp(- (x - µ) 2 / 2 σ 2 ) dx + ½ = ½ M σ\2π DIT 111 PROBABILITY AND QUEUEING THEORY 82 NOTES Anna University Chennai µ I f(x) dx = 0 M M = µ Therefore median = µ 1.8.3.7 Mode of the normal distribution N (μ, σ) Mode of a continuous random variable X is defined as the values of x for which the density function f(x) is maximum. For the normal distribution N (μ, σ) · f(x) = 1 I x exp(- (x - µ) 2 / 2 σ 2 ) dx σ\2π -· log f(x) = k - 1 (x - µ) 2 2 σ 2 Differentiating with respect to x, f '(x) = - (x - µ) f (x) σ 2 f '(x) = - (x - µ) f(x) σ 2 = 0 when x = µ f " (x) = -{(x - µ) f ‘(x) + f(x)} σ 2 f " (x) = - f(µ) <0 x = μ σ 2 Therefore f(x) is maximum at x = µ. Therefore mode is µ Note: Have you noticed anything while finding the mean, median and mode? Yes! For normal distribution mean, median and mode are same 1.8.3.8 Central moments of the normal distribution N (µ r , σ) Central moments µ r of N (μ, σ) are given by μ r = E( x - µ) r · = 1 I (x - µ) r exp(- (x - µ) 2 / 2 σ 2 ) dx σ\2π -· · = 1 I (\2 σt) r exp (-t 2 )dt \π -· DIT 111 PROBABILITY AND QUEUEING THEORY 83 NOTES Anna University Chennai · = 2 r / 2 σ r I t r exp (-t 2 )dt vπ -· Case i) r is an odd integer, that is r = 2n + 1 · Therefore µ 2n+1 = 2 (2n +1)/2 σ 2n +1 I t 2n+1 exp (-t 2 )dt \π -· = 0, (since the integrand is an odd function) Case ii) r is an even integer, i.e. r = 2n · µ 2n = 2 n σ 2n I t 2n exp(-t 2 ) dt \π -· · = 2 n σ 2n 2 I t 2n exp(-t 2 ) dt (since t 2n exp(-t 2 ) is an even function of t) \π 0 (put t 2 = u; 2tdt = du) · = 2 n σ 2n I u n- ½ e -u du \π 0 = 2 n σ 2n |( n+ ½) (1) \π = 2 n σ 2n (2n -1 ) ( 2n - 1) \π 2 2 = 2 n σ 2n (2n -1 ) ( 2n – 3) ( 2n - 3) \π 2 2 2 = 2 n σ 2n (2n -1 ) ( 2n – 3) - - - - 1 | ½ \π 2 2 2 ( There are n terms in the series 1 . 3. 5.. . . . .(2n - 3)(2n - 1) ) For example consider 1.3.5.7 we have 4 terms i.e. 7 -1 + 1 i.e in general (n - 1)/2 + 1. 2 Therefore in the series 1 . 3. 5.. . . . .(2n - 3)(2n - 1) ) there are 2n – 1 – 1 + 1 = n) 2 = 2 n σ 2n 1.3.5. ...... (2n - 1) \π ( | ½ = \π) \π 2 n µ 2n = σ 2n 1.3.5. ...... (2n - 1) Form (1) we get µ 2n-2 = 2 n - 1 σ 2n - 2 |( n - ½) (2) \π DIT 111 PROBABILITY AND QUEUEING THEORY 84 NOTES Anna University Chennai From (1) and (2) we get µ 2n = 2 σ 2 (n – ½) µ 2n-2 µ 2n = 2 σ 2 (n – ½) µ 2n-2 which gives a recurrence relation for the even order central moments of the normal distribution. 1.8.3.9 Mean deviation about the mean of the normal distribution The central moment of the first order of a random variable X is called the mean deviation (MD) about the mean X .i.e. E{| X –E(X)| } For the normal distribution N(μ, σ) · MD = I |x - µ| 1 exp(-(x - µ ) 2 /2σ 2 ) dx -· σ \2π · = 1 I \2σ t | exp(-t 2 ) \2σ dt σ \2π -· · = \2 σ I |t| exp(-t 2 ) dt \π -· · = 2\2 σ I |t| exp(-t 2 ) dt (since |t| exp(-t 2 ) is an even function. ) \π 0 · = \2 σ (exp(-t 2 )) \π 0 = \2 σ \π 1.8.3.10 Moment generating function of N(μ, σ) and N(0,1) The moment generating function of N(μ, σ) Mx(t) = M σZ + µ (t) = e µt µ Z (σ t) = e µt exp(σ 2 t 2 /2) = exp[t(µ + σ 2 t / 2 )] DIT 111 PROBABILITY AND QUEUEING THEORY 85 NOTES Anna University Chennai Now Mx(t) = 1 + t (µ + σ 2 t / 2 ) + t 2 (µ + σ 2 t /2 ) 2 + t 3 (µ + σ 2 t / 2 ) 3 + - - - - - 1! 2! 3! E(X) = Coefficient of t / 1! = µ E(X 2 ) = Coefficient of t 2 / 2! = σ 2 + µ 2 E(X 3 ) = Coefficient of t 3 / 3! = 3 μ σ 2 + µ 3 and so on. Moment generating function of N(0,1) Mz(t) = E(e tz ) · = I e tz φ(z)dz -· · = I 1 exp(-z 2 /2) e tz dz -· \2π · = I 1 [exp-(z 2 – 2tz)/2] dz -· \2π · = I 1 [exp-[(z-t) 2 – t 2 )/2]] dz -· \2π Put u = z – t ; du = dz/\2 \2 · = 1 exp(t 2 /2)\2 I exp(-u 2 )du \2π -· = 1 exp(t 2 /2) |½ ( |½ = \π) \π = exp(t 2 /2) Note: If X has the distribution N(μ,σ) then Y = aX + b has the distribution N(a μ + b, a σ) with mgf M Y (t) = exp{[(aµ + b)t] + [(a 2 σ 2 )t 2 /2]} 1.8.3.11 Additive property of normal distribution If Xi (i = 1,2 .. n) be n independent normal random variables with mean µ i and variance σ i 2 then DIT 111 PROBABILITY AND QUEUEING THEORY 86 NOTES Anna University Chennai n n n a i x i is also a normal random variable with mean a i µ i and variance a i σ i 2 . i = 1 i = 1 i = 1 M n (t) = M a1x1 (t). M a2x2 (t) . - - - - - M an xn (t) aixi i = 1 = exp( a 1 µ 1 t) exp(a 1 2 σ 1 2 t 2 /2) x exp( a 2 µ 2 t) exp(a 2 2 σ 2 2 t 2 /2) x - - - - x exp( a n µ n t) exp(a n 2 σ n 2 t 2 /2) x = exp[( a i µ i )t} exp( a i 2 σ i 2 t 2 /2 ) which is the mgf of normal random variable with mean a i µ i and variance a i 2 σ i 2. Note: Putting a 1 = a 2 = 1 and a 3 = a 4 = - - - - = an = 0, we get the following result, in particular: If X 1 is N(µ 1 , σ 1 ) and X 2 is N(µ 2 , σ 2 ), then X 1 + X 2 is N(µ 1 + µ 2, \ σ 1 2 + σ 2 2 ). Similarly X 1 - X 2 is N(µ 1 - µ 2, \ σ 1 2 + σ 2 2 ). 1.8.3.12Normal distribution as limiting form of binomial distribution. When n is very large and neither p and q is very small, the standard normal distribution can be regarded as the limiting from of the standardised binomial . When X follows the binomial distribution B(n,p), the standrdised binomial variable Z is given by Z = (X – np) /\npq with step size1, Z varies from –np/ \npq to np/ \npq with step size 1/\npq . When neither p nor q is very small and n is very large, Z varies from - 8 to 8 with infinite small step size. Hence, in the limit, the distribution of Z may be expected to be a continuous distribution extending from - 8 to 8 , having mean 0 and SD 1. If X follows B(n ,p), then mgf of X is given by M X (t) = (q + p e t ) n If Z = (X – np) /\npq then M Z (t) = e (-npt / \npq ) {q + pe t/ \npq } log M Z (t) = - npt + n log {q + pe t/ \npq } \npq = - npt + nlog q + p 1 + t + t 2 + t 3 + - - - - - \npq \npq 2npq 6 (npq) 3/2 = - npt + nlog 1 + p t + p t 2 + p t 3 + - - - - - \npq \npq 2npq 6 (npq) 3/2 { } { } [ ] [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 87 NOTES Anna University Chennai = - npt + n p t 1 + t + t 2 + - - - - \npq \npq 2\npq 6 n 2 p 2 q 2 - 1 p 2 t 2 1 + t + t 2 + 2 + - - 2 npq 2\npq 6n 2 p 2 q 2 = t 2 / 2 + terms containing 1/\n and lower powers of n. lim log M Z (t) = t 2 /2 n  · log e [ lim M Z (t) ] = t 2 /2 n  · lim M Z (t) = exp(t 2 /2) n  · which is the mgf of the standard normal distribution. Hence the limit of the standardized binomial distribution, as n  ·, is the standardized normal distribution. Areas of application: Normal distribution is the most important continuous probability distribution in the field of statistics. It describes many phenomena that occur in industry , in error calculations of experiments, statistical quality control, in nature like rainfall and meteorological studies etc. Example1: The marks obtained by a number of students in a certain subject are assumed to be approximately normally distributed with mean 55 and SD(standard deviation) 5. If 5 students are taken at random from this set, what is the probability that 3 of them would have scored marks above 60? Solution: If X represents the marks obtained by the students, X follows the distribution N(55,5). P(a student scores above 60) = P(X >60) = P(60 < X < ·) = P[(60 – μ)/ σ < (X – μ)/ σ < ·] = P[(60 – 55)/5 < Z < ·](where Z is the standard normal variate.) [ ] { } { } ] DIT 111 PROBABILITY AND QUEUEING THEORY 88 NOTES Anna University Chennai = P[1 < Z < ·] In the problem we are converting to standard normal distribution N(0,1) i.,e mean μ = 0 and σ = 1.Now to find the values we will be using the normal distribution table .it will be better if you draw the normal curve to find the area. Remember the the total area under the normal curve is 1. Area to the left of µ = 0 is ½ and to the right is ½. P[1 < Z < ·] = 0.5 – P(0 < Z < 1) =0.5 - .3413 (from the table of areas) = .1587 Therefore P(a student scores above 60) = 0.1587 Let p = P(a student scores above 60) = 0.1587 q = .8413 Since p is the same for all the students, the number Y, of students scoring above 60 follows a binomial distribution. P( 3 of the students scoring above 60 among the five students) = nCxp x q n-x = 5C 3 (0.1587) 3 (0.8413) 2 = 0.0283 Example 2: The mean and SD of a certain group of 1000 high school grades, that are normally distributed are 78% and 11% respectively. Find how many grades were above 90%? Solution: Let X represents the no: of grades. Given μ = 78% = 0.78 & σ = 11% = 0.11 P(X>90%) =P(X>0.09) = P(0.09 < X < ·]) = P[(0.09 – 0.78)/ 0.11 < (X – μ)/ σ < ·] = P[1.090 < Z < ·] DIT 111 PROBABILITY AND QUEUEING THEORY 89 NOTES Anna University Chennai = 0.5 –P( 0< Z < 1.090) = 0.5 – 0.3621 = 0.1379 Therefore out of 1000 high school grades the no : of grades above 90% will be 0.1379 x 1000 = 137.9 [ 138 Example 3: The local authorities in a certain city install 10,000 electric lamps in the streets of the city. If these lamps have an average life of 1000 burning hours with a SD of 200h, how many lamps might be expected to fail i) in the first 800 burning hours?ii) between 800 and 1200 burning hours. After how many burning hours would you expect iii)10% of the lamps to fail? iv) 10% of the lamps to be still burning? Assume that the life of lamps is normally distributed. Solution: If X represents life of the electric lamps , X follows the distribution N(1000, 200) Total no: of electric lamps in the city, N = 10,000 i)P(X < 800) =P(-· <X < 800) = P(-· < (X – μ)/ σ < (800 – 1000)/200 ) = P(-· < Z < -1) DIT 111 PROBABILITY AND QUEUEING THEORY 90 NOTES Anna University Chennai = 0.5 - P(-1< Z< 0) = 0.5 – P(0 < Z <1) (by symmetry) = 0.5 – 0.3413 = 0.1587 The number of lamps expected to fail in the first 800 burning hours out of 10,000 electric lamps are 10,000 x 0.1587 = 1587. ii)P(800<x<1200) = P[(800 – 1000)/200 < (X – μ)/ σ < (1200 – 1000)/200 ] = P ( -1 < Z < 1) = 2 x P(0 < Z < 1) = 2 x 0.3413 = 0.6826 The number of lamps expected to fail in between 800 and 1200 burning hours out of 10,000 electric lamps are 10,000 x 0.6826 = 6826. iii) P(X< x 1 ) = 0.1, we have to find x 1 =>P(-·< X < x 1 ) = P(-· < (X – μ)/ σ < (x 1 - 1000)/200 ) = 0.1 =>P(-· < Z < (x 1 - 1000)/200 ) = 0.1 Let a = (x 1 - 1000)/200) From -· to a the area is given as 0.1, but we know that -· to 0 the area is 0.5. therefore a will be lying in the left half of the normal curve. =>P(-· < Z < (x 1 - 1000)/200 ) = 0.1 =>0.5 – P((x 1 - 1000)/200) < Z < 0) = 0.1 =>0.5 – P(0< Z < (1000 - x 1 )/200 ) = 0.1 => P(0< Z < (1000 - x 1 )/200)) = 0.4 (search for the value nearer to 0.4 inside the table ) Therefore (1000 - x 1 )/200) = 1.29 x 1 = 744 DIT 111 PROBABILITY AND QUEUEING THEORY 91 NOTES Anna University Chennai iv) P(X>x 2 ) = 0.1 => P(x 2 < X < 8)= 0.1 =>P[(x 2 – 1000)/200 < (X – μ)/ σ < ·] = 0.1 =>P[(x 2 – 1000)/200 < Z < ·] = 0.1 Let (x 2 – 1000)/200 = b => 0.5 – P(0 < Z < (x 2 – 1000)/200 ) = 0.1 P(0 < Z < (x 2 – 1000)/200 ) = 0.4 Therefore ( x 2 - 1000 )/200) = 1.29 x 2 = 1256 Example 4: The marks obtained by the students in Mathematics, Physics and Chemistry in an examination are normally distributed with the means 52, 50 and 48 and with standard deviations 10,8 and 6 respectively. Find the probability that a student selected at random has secured a total of i) 810 or above and ii) 135 or less. Solution: Let X,YZ Denote the marks obtained by students in mathematics, physics and chemistry respectively. X follows N(52,10), Y follows N(50,8) and Z follows N(48,6) By the additive property of normal distribution U = X + Y+ Z follows the distribution N{52 + 50 + 48, \10 2 + 8 2 + 6 2 } i.e. N(150, 14.14) i) P(U=180) = P{ (180 -150/14.14 < (U-150)/14.14 < · } = P {2.12 < Z < ·} = 0.5 – P(0 < Z < 2.12) = 0.5 – 0.4830 = 0.0170 ii) P(U = 135) = P{-· < U < 135} = P(-· < (U-150)/14.14 < (135 - 150)/14.14 } = P(-· < Z < -1.06) = 0.5 – P(-1.06< Z < 0) DIT 111 PROBABILITY AND QUEUEING THEORY 92 NOTES Anna University Chennai = 0.5 – P(0 < Z <1.06) = 0.5 – 0.3554 = 0.1446 Example 5: In a normal population with mean 15 and SD 3.5, it is found that 647 observations exceeds 16.25. What is the total number of observations in the population? Solution: Let N be the no: of observation. Given that N x P(X> 16.25) = 647 (1) P(X> 16.25) = P(16.25 < X < ·) = P(16.25 -15/3.5 < (X – μ)/ σ < · ) = P(0.3571 < Z < ·) = 0.5 – P(0 < Z < 0.3571) = 0.5 – 0.1406 = 0.3594 Therefore P(X> 16.25) = 0.3594 Substituting in (1) N x 0.3594 = 647 N = 647 / 0.3594 = 1800 Example 6: In a normal distribution, 7% of the items arte under 35 and 89% are under 63. what are the mean and standard deviation of the distribution? What percentage of the items are under 49? Solution: P(X < 35) = 0.07 P(X <63) = 0.89 P(-· <X<35) = 0.07 P(-· <X<63) = 0.89 P(-· <(X – μ)/ σ < (35- μ)/ σ ) = 0.07 P(-· <(X– μ)/ σ < (63– μ)/ σ) = 0.89 P(-· < Z < (35- μ)/ σ ) = 0.07 P(-· < Z < (63 – μ)/ σ) = 0.89 {Let (35- μ)/ σ = a and (63 – μ)/ σ = b.} DIT 111 PROBABILITY AND QUEUEING THEORY 93 NOTES Anna University Chennai 0.5 – P((35- μ)/ σ <Z< 0 ) = 0.07 0.5 + P((63- μ)/ σ < Z< 0 ) = 0.89 0.5 – P(0 < Z < (μ -35)/ σ ) = 0.07 0.5 + P((63- μ)/ σ < Z< 0 ) = 0.89 P(0 < Z < (μ -35)/ σ ) = 0.43 P((63- μ)/ σ < Z< 0 ) = 0.39 (μ -35)/ σ = 1.47 (1) (63- μ)/ σ = 1.23 (2) From 1 & 2 35 – μ = -1.47 σ 63 – μ = 1.23 σ Solving these two equations we will get μ = 50.24 σ = 10.37 Therefore mean = 50.24 and SD = 10.37 P (X <49) = P(-· <X< 49) = P(-· <(X – μ)/ σ < (49 -50.24)/10.37) = P(-· < Z < –0.12) = 0.5 – P(–0.12 < Z < 0) = 0.5 – P(0 < Z < 0.12) = 0.5 – 0.0478 = 0.4522 = 45.22% P (X <49) = 45.22% Example 7: There are 400 students in the first year class of an engineering college. The probability that any student requires a copy of a particular mathematics book from the college library on any day is 0.1. How many copies of the book should be kept in t he library so that the probability may be greater than 0.95 that one of the students requiring a copy from the library has to come back disappointed? (Use normal approximation to the binomial distribution) Solution: p = P(a student require book from) = 0.1 q = 0.9, n = 400 If X represents the number of students requiring the book, then X follows a binomial distribution with mean = np = 400 x 0.1 = 40 and SD = \npq = 6 In the problem it is given that X follows the distribution N(40,6) Let x 1 be the required number of books, satisfying the given condition i.e P(X < x 1 ) >0.95 P{-· < X < x 1 }>0.95 => P(-· < (X-40)/6 < (x 1 - 40)/6 ) > 0.95 DIT 111 PROBABILITY AND QUEUEING THEORY 94 NOTES Anna University Chennai => 0.5 + P(0 < Z < (x 1 - 40)/6) > 0.95 => P(0 < Z < (x 1 - 40)/6) ) > 0.45 From the table of areas under normal curve, we find that P(0 < Z < 1.65) > 0.45 Therefore (x 1 - 40)/6 = 1.65  m = 49.9 [ 50  Therefore atleast 50 copies of the book should be kept in the library. Example 8: If X and Y are independent random variables following N(8,2) and N(12, 4\3) respectively find the value of λ such that P(2X – Y , 2λ) = P(X +2Y ¡ λ) Solution: Given X follows N(8,2) and Y follows N(12, 4\3) Let A = 2X – Y and B = X + 2Y If X 1 is N(µ 1 , σ 1 ) and X 2 is N(µ 2 , σ 2 ) then X 1 - X 2 is N(µ 1 - µ 2, \ σ 1 2 + σ 2 2 ). (By additive property) 2X – Y is N( 2x8 – 12, \ 4 x 4 + 1 x 48 [since E(aX) = a E(X), Var(aX) = a 2 Var(X)]] i.e.A follows N(4,8) X + 2Y is N( 8 + 2 x 12, \4x4 x 48) i.e B follows N(32,14) Given P(2X – Y , 2λ) = P(X +2Y ¡ λ) => P(A , 2λ) = P(B ¡ λ) => P[(A-4)/8 , (2λ – 4)/8]= P[(B- 32)/14 ¡ (λ – 32)/14)] => P[ Z , (2λ – 4)/8] = P[ Z ¡ (λ – 32)/14)] => P[-· , (2λ – 4)/8] = P[(λ – 32)/14) , 8 ] DIT 111 PROBABILITY AND QUEUEING THEORY 95 NOTES Anna University Chennai => P[-· , (2λ – 4)/8] = P[-· , - (λ – 32)/14) ] (bu symmetry) => (2λ – 4)/8 = - (λ – 32)/14 Therefore λ = 8.67 Normal Deviate .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 z 0.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359 0.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753 0.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141 0.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517 0.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879 0.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224 0.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549 0.7 .2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852 0.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133 0.9 .3159 .3186 .3212 .3238 .3264 3289 .3315 .3340 .3365 .3389 1.0 .3413 .3438 .3461 .3485 .3508 .3531 .3554 .3577 .3599 .3621 1.1 .3643 .3665 .3686 .3708 .3729 .3749 .3770 .3790 .3810 .3830 1.2 .3849 .3869 .3888 .3907 .3925 .3944 .3962 .3980 .3997 .4015 1.3 .4032 .4049 .4066 .4082 .4099 .4115 .4131 .4147 .4162 .4177 1.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319 1.5 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441 1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545 1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633 1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706 1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767 2.0 .4772 .4778 .4783 .4788 .4793 .4798 .4803 .4808 .4812 .4817 2.1 .4821 .4826 .4830 .4834 .4838 .4842 .4846 .4850 .4854 .4857 2.2 .4861 .4864 .4868 .4871 .4875 .4878 .4881 .4884 .4887 .4890 2.3 .4893 .4896 .4898 .4901 .4904 .4906 .4909 .4911 .4913 .4916 2.4 .4918 .4920 .4922 .4925 .4927 .4929 .4931 .4932 .4934 .4936 2.5 .4938 .4940 .4941 .4943 .4945 .4946 .4948 .4949 .4851 .4952 2.6 .4953 .4955 .4956 .4957 .4959 .4960 .4961 .4962 .4963 .4964 2.7 .4965 .4966 .4967 .4968 .4969 .4970 .4971 .4982 .4973 .4984 2.8 .4974 .4975 .4976 .4977 .4977 .4978 .4979 .4979 .4980 .4981 2.9 .4981 .4982 .4982 .4983 .4984 .4984 .4985 .4985 .4986 .4986 3.0 .4865 .4987 .4987 .4988 .4988 .4989 .4989 .4989 .4990 .4990 4.0 .4997 DIT 111 PROBABILITY AND QUEUEING THEORY 96 NOTES Anna University Chennai Have you understood ? Say true or false. Justify your answer. 1.Exponential random variable is not memoryless. 2.Mean of uniform distribution is (b – a) /2. 3.Mean, median and mode of normal distribution are same. (Answers:1.False,2.False,3.True) Short answer questions. 1.Waht do you mean by memoryless property of exponential distribution? 2.Discuss uniform distribution and explain its mean and variance. 3.State the reproductive property of normal distribution. 4.why is normal distribution considered as an important distribution. 5.Write down the mgf of exponential distribution and hence derive its mean. Try yourself ! 1 Electric trains on a certain line run every half hour between mid-night and six in morning, What is the probability that a man entering that station at a random time during this period will have to wait atleast twenty minutes. (Solution: 1/3) 2. If X has uniform distribution in(-a, a), a>0 find a such that P(|X| < 1) = P(|X| >1) (Solution: a = 2) 3. The length of the shower in a tropical island in rainy season has an exponential distribution with parameter 2, time being measured in minutes. What is the probability that it will last for atleast one more minute? (Solution: 0.1353 ) 4. The marks obtained by a number of students in a certain subject are approximately normally distributed with mean 65 and standard deviation 5. If 3 students are selected at random from this group, what is the probability that atleast 1 of them would have scored above 75. (Solution: 0.0667) 5. In an examination, a student is considered to have failed, secured second class, first class and distinction, according as he scores less than 45%, between 45% and 60%, between 60% and 75% and above 75% respectively. In a particular year 10% of the students failed in examination and 5% of the students get distinction. Find the percentage of students who have got first class and second class. (Assume normal distribution of marks) (Solution: 38, 47s) DIT 111 PROBABILITY AND QUEUEING THEORY 97 NOTES Anna University Chennai 1.9 FUNCTIONS OF RANDOM VARIABLES 1.9.1 Transformation of one dimensional random variable Let X be a continous random variable with pdf f x ( X ). We determine the density function of a new random variable Y = g (x ) where g is a function. Further let y = g (x) be strictly monotonic function of x. Now f Y (y) the pdf of Y is given f Y (y) = f X (x) | dx | or f X (x) F 1 (x) = f (x) | dy | | g 1 (x)| g(x) is a strictly increasing function of x case i : g (x) is a strictly increasing function of x. F Y (y) = P ( Y , y), where F Y (y) is the cdf of Y = P[g(x) , y] = P[X , g -1 (y)] = Fx(g -1 (y)) Differentiating on both sides with respect toy, f Y (y) = f X (x) dx where x = g -1 (y) (1) dy case ii) g(x) is a strictly decreasing function of x. F Y (y) = P ( Y , y), where F Y (y) is the cdf of Y = P[g(x) , y] = P[X ¡ g -1 (y)] = 1 - P[X , g -1 (y)] = 1 - Fx(g -1 (y)) f Y (y) = –f X (x) dx (2) dy combining (1) and(2), we get f Y (y) = f X (x) | dx | = f X (x) | dy | | g 1 (x)| When x = g -1 (y) takes finitely many values x 1 ,x 2 ,x 3 , - - x n, then f Y (y) = f X (x 1 ) | dx 1 | + f X (x 2 ) | dx 2 | + - - - - -+ f X (x n ) | dx n | | dy | | dy | | dy | Example 1: Consider a random variable X with probability density function f X (x) = e -x , x=0 with the transformation y = e -x . Find the transformed density function? DIT 111 PROBABILITY AND QUEUEING THEORY 98 NOTES Anna University Chennai Solution: f Y (y) = f X (x) | dx | | dy | Given f X (x) = e -x and y = e -x => dy = – e -x dx Therefore f Y (y) = e -x | 1 | = 1 |– e -x | f Y (y) = 1, 0< y , 1 = 0, otherwise. Example: If X is uniformly distributed in the interval ( –π/ 2 , π/ 2 ), find the pdf of Y = tanX ? Solution: As X exists in the range ( –π/ 2 , π/ 2 ), Y = tan X exists in the range –· to ·. As X is uniformly distributed its pdf is given by f(x) = 1/(b - a) a < x < b 0, otherwise There f X (x) = 1/π Y = tanX => dy/dx = sec 2 x f Y (y) = f X (x) | dx | | dy | = 1/ π (1 / sec 2 x ) = (1/ π) (1/1 + y 2 ) , –·<y< ·. [ as sec 2 x – tan 2 x = 1 => sec 2 x = 1 + tan 2 x] Example 2: The random variable Y is defined as ½ (X + |X|), where X is a random variable. Find the density and distribution function of Y ? Solution: Given Y = ½ (X + |X| ) y = 0 when x < 0 y = x when x ¡ 0 Now, distribution function F Y (y) = P(Y , y) For y < 0, F Y (y) = 0 For y ¡ 0 , F Y (y) = P(Y , y) = P[(X , y)/ X¡ 0] = P( 0 , X , y) = P( 0 , X , y) P(X ¡ 0) 1 - P(X < 0) DIT 111 PROBABILITY AND QUEUEING THEORY 99 NOTES Anna University Chennai = Fx(y) – Fx(0) 1 - Fx(0) To find the density function, F 1 (x) = f (x) So f Y (y) = fx(y) – Fx(0) 1 - Fx(0) Example 3: If Y = x 2 , where X is a Gaussian random variable with zero mean and variance σ 2 , find the pdf of the random variable Y. Solution: F Y (y) = P( Y , y) =P(X 2 , y) = P(–\y , X , \y), if y ¡ 0 = Fx(\y) – Fx(–\y) Differentiating with respect to y, f Y (y) = 1 .{f X (\y) + f X (–\y) }, if y ¡ 0 (1) 2\y = 0, if y < 0. It is given that X follows N(0,σ) Therefore f X (x) = 1 . exp(-x 2 / 2σ 2 ), –·<x< · σ\2π using this value in (1) , we get f Y (y) = 1 . exp(-y / 2σ 2 ), y >0 σ\2πy Example 4: Find the density function of Y = aX + b in terms of the density function of X? Solution i)Let a > 0 F Y (y) = P(Y, y) = P(aX + b , y ) = P X , ( y – b ) (since a>0) a = F X y – b (1) a ii)Let a< 0 F Y (y ) = P(Y, y) = P(aX + b , y ) = P(aX , y - b) = P X ¡ y - b a = 1 - P X < ( y – b ) a = 1 - F X y – b (2) a DIT 111 PROBABILITY AND QUEUEING THEORY 100 NOTES Anna University Chennai from (1) f Y (y) = 1 f X y – b (3) a a from (2) f Y (y) = – 1 f X y – b (4) a a combining (3) & (4) f Y (y) = 1 f X y – b |a| a Try yourself ! 1.If Y = X 2 find the pdf of Y if X has pdf f X (x) = 1/3, -1 < x < 2 = 0, elsewhere (Solution: f Y (y) = 1/(3\y), 0 < y< 1 1/(6\y), 1 < y< 4 0, otherwise 2.If the density function of a continuous random variable X is given by f X (x) = 2/9 (x +1), for -1< x < 2, and = 0, otherwise find the density function of Y = ½(X + |X|) Solution: f Y (y) = (y +1) /4, 0<y<2 0, otherwise REFERENCES: 1. T.Veerarajan, "Probability, statistics and Random Process", Tata McGraw Hill, 2002. 2. P.Kandasamy, K. Thilagavathi and K. Gunavathi, " Probability, Random Vari- ables and Random processors", S. Chand, 2003. DIT 111 PROBABILITY AND QUEUEING THEORY 101 NOTES Anna University Chennai UNIT 2 TWO DIMENSIONAL RANDOM VARIABLES - Introduction - Two dimensional random variables - Marginal and conditional probability - Expectation - Covariance - Correlation - Regression - Transformation of random variables 2.1 INTRODUCTION In the last unit we have considered one dimensional random variable. But in many practical problems several random variables interact with each other and we are interested in the joint behavior of these random variables 2.2 LEARNING OBJECTIVES - Knowledge of basic concepts and results relating to random variables and their applications. - Familiarity with some of the distributions commonly used to represent real-life situations. - Knowledge of a variety of parametric and nonparametric statistical methods. - Appreciation of the theoretical foundations of statistical methods. - Enhancement of mathematical skills which are particularly relevant to statistics. DIT 111 PROBABILITY AND QUEUEING THEORY 102 NOTES Anna University Chennai 2.3 TWO DIMENSIONAL RANDOM VARIABLES: 2.3.1 Definition: Let S be the sample space of a random experiment. Let X and Y be two random variables defined on S. Then the pair ( X,Y ) is called a two dimensional random variable or a bivariate random variable. 2.3.2 Probability Function of ( X, Y ) If (X,Y) is a two dimensional discrete random variable such that P(X = x i , Y = y j ) = p ij , then p ij is called the probability mass function or simply the probability function of (X,Y ) provided the following conditions are satisfied. i) p ij = 0, for all i and j ii) Σ Σ p ij = 1 j i The set of triplets { x i , y j , p ij }, I = 1,2,……m, …… j = 1,2,…..n,…., is called the joint probability distribution of ( X,Y ) Joint Probability Density Function If ( X, Y ) is a two dimensional continous random variable such that P x – dx = X = x + dx and y – dy = Y = y + dy = f(x,y) dx dy, then f(x,y) is called 2 2 2 2 the joint pdf of (X,Y), provided f(x,y) satisfies the following conditions i) f(x,y) ¡ 0, for all (x,y) ε R, where R is the range space ii) IIf(x,y) dx dy = 1 R Moreover if D is a subspace of the range space R, P {(X,Y) ε D} is defined as P{(X,Y) ε D } = II f(x,y) dx dy. In particular D d b P {a , X , b, c , Y , d }= II f (x,y) dx dy c a 2.3.3 Cumulative Distribution Function (cdf) If (X,Y ) is a two dimensional random variable ( discrete or continous ), then F ( x,y) = P {X , x and Y , y } is called the cdf of ( X,Y) DIT 111 PROBABILITY AND QUEUEING THEORY 103 NOTES Anna University Chennai In the discrete case, F(x,y) = Σ Σ p ij j i In the continuous case, y x F(x,y) = II f ( x,y) ds dy -· -· 2.3.3.1 Properties of F (x,y ) F (-·, y ) = 0 = F (x, -· ) and F (·,· ) = 1 P { a < X < b, Y = y }= F (b, y ) – F ( a, y ) P { X , x, c < Y < d}= F(x,d) – F (x,c) P (a < X < b, c < Y <d} = F(b,d) – F ( a,d) – F ( b,c) + F (a,c) At points of continuity of f(x,y ) c 2 F = f(x,y) cxcy 2.4 A) MARGINAL PROBABILITY DISTRIBUTION P ( X = x i ) = P { ( X=x i and Y = y i ) or ( X = x i and Y = y 2 ) or etc } = p i1 + p i2 + …… = Σ p ij j P ( X=x i ) = Σ p ij is called the marginal probability function of X. It is defined for X = x 1 , j x 2 ,…… and denoted as P i*. The collection of pairs { x i , p i* }, i = 1,2,3,…..is called the marginal probability distribution of X. Similarly the collection of pairs { y j , p *j }, j = 1,2,3,……is called the marginal probability distribution of Y, where p *j = Σ p ij = P(Y = y i ) i In the continuous case P { x – 1 dx , X , x + 1 dx }, -· < Y < · 2 2 x + 1 dx · 2 = I I f(x,y )dx dy -· x-1 dx 2 · = I f(x,y) dy dx [ since f(x,y) may be treated as a constant in ( x-1dx , x + 1 dx)] - · 2 2 = f X (x) dx, say DIT 111 PROBABILITY AND QUEUEING THEORY 104 NOTES Anna University Chennai · f X (x) = I f(x,y)dy is called the marginal density of X. - · · Similarly, f Y (y) = I f(x,y)dx is called the marginal density of Y. - · Note: P (a , X , b ) = P(a , X , b, - · < Y < · ) · b = I I f (x,y)dxdy - · a b · b = I I f (x,y)dy dx = I f X (x) dx a - · a d Similarly, P( c , Y , d) = I f Y (y) dy c 2.4 B)CONDITIONAL PROBABILITY DISTRIBUTION : P X=x i = P{X=x i Y=y j } = p ij is called the conditional probability function of Y=y j P{Y=y j } p *j X given that Y = y j* The collection of pairs, { x i , p ij } i = 1,2,3….. p *j is called the conditional probability distribution of X, given Y = y j* Similarly the collection of pairs { Y j , p ij } j = 1,2,3….. is called the conditional p i* probability distribution of Y, given X = x j*. In the continuous case, P { x – 1 dx , X < x + 1 dx / Y = y } 2 2 = P { x – 1 dx , X , x + 1 dx y – 1 dy , Y , y + 1 dy } 2 2 2 2 = f(x,y ) dx dy = { f(x,y) } dx f Y (y) dy f Y (y) f(x,y) is called the conditional density of X, given Y, and is denoted by f x f Y (y) y [ ] ¯ .' DIT 111 PROBABILITY AND QUEUEING THEORY 105 NOTES Anna University Chennai Similarly f(x,y) is called the conditional density of Y, given X, and is denoted by f y F X (x) x Independent Random Variables If (X,Y) is a two dimensional discrete random variable such that P { X = x i / Y = y j } = P(X=x i ) i.e p ij = p i* , i.e., p ij = p i* x p *j for all i.j then X and Y are said to be p *j independent. Similarly If (X,Y) is a two dimensional continuous random variable such that f(x,y) = f X (x) X f Y (y), then X and Y are said to be independent random variable. Example 1: If X denotes the number of aces and Y the number of queens obtained when 2 cards are drawn at random (without replacement) from a deck of cards, obtain the joint probability distribution of (X,Y) Solution: Let X denote the number of aces and Y denote the number of queens. There are 4 ace cards and 4 queen cards in a deck. We are taking 2 cards from a deck of 52 cards. So X can take the values 0, 1, 2 and Y can take 0, 1, 2. The joint probability distribution of (X,Y) is found as follows. X Y 0 1 2 0 946/1326 176/1326 6/1326 1 176/1326 16/1326 0 2 6/1326 0 0 P(X = 0, Y =0) = P(drawing 2 cards none of which is a ace or queen) = P(drawing 2 cards from the rest of 44 (52 - 8) cards) = 44C 2 = 946 . 52C 2 1326 P(X =0, Y =1) = 4C 1 x 44C 1 = 176 52C 2 1326 P(X = 0, Y = 2) = 4C 2 = 6 . 52C 2 1326 ¯ .' DIT 111 PROBABILITY AND QUEUEING THEORY 106 NOTES Anna University Chennai P(X = 1, Y= 0) = 4C 1 x 44C 1 = 176 52C 2 1326 P( X = 1, Y = 1) = 4C 1 x 4C 1 = 16 . 52C 2 1326 P(X = 1, Y = 2) = 0 (since only 2 cards are only drawn ) P(X = 2, Y = 0) = 4C 2 = 6 . 52C 2 1326 P(X = 2, Y = 1 ) = 0 P(X = 2, Y = 2 ) = 0 Sum of all the cell probabilities = 946/1326 + 2[176/1326] + 2[6/1326] + 16/1326 = 1 Example 2 : The joint distribution of X 1 and X 2 is given by f(x) = x 1 + x 2 , x 1 = 1,2 and 3, x 2 = 1,2. Find a) the marginal distribution of X 1 andX 2 . 21 b) conditional distribution of X 1 given X 2 = 2 and X 2 given X 1 = 1.Are X 1 and X 2 are independent? Also find the probability distribution of X + Y? Solution: The joint probability distribution of(X,Y) is given below X 1 X 2 1 2 3 1 2/21 3/21 4/21 2 3/21 4/21 5/21 a)Marginal probability distribution of X 1 X 1 =i 2 p i * = Σ p i j j =1 1 P 11 + P 12 = 2/21 + 3/21 = 5/21 2 P 21 + P 22 = 3/21 + 4/21 = 7/21 3 P31 + P32 = 4/21 + 5/21 = 9/21 DIT 111 PROBABILITY AND QUEUEING THEORY 107 NOTES Anna University Chennai Marginal probability distribution of X 2 X 2 = j 3 p * j = Σ p i j i =1 1 P 11 + P 21 + P 31 = 2/21 + 3/21 + 4/21 = 9/21 2 P 12 + P 22 + P 32 = 3/21 + 4/21 + 5/21 =12/21 If X 1 and X 2 are independent we will have p i* x p *j = p ij Let i = 2 and j = 1, P 2* x P *1 = 7/21 x 9/21 = 63/21 ? P 21 (= 3/21) Therefore X 1 and X 2 are not independent. b) Conditional distribution of X 1 given X 2 = 2 X 1 = i p i 2 p * 2 1 P 12 3/21 P 12 + P 22 + P 32 = 3/21 + 4/21 + 5/21 = 1/4 2 P 22 4/21 P 12 + P 22 + P 32 = 3/21 + 4/21 + 5/21 = 1/3 3 P 32 5/21 P 12 + P 22 + P 32 = 3/21 + 4/21 + 5/21 = 5/12 Conditional distribution of X 2 given X 1 = 1 X 2 = j p 1j p 1 * 1 P 11 2/21 P 11 + P 12 2/21 + 3/21 = 2/5 2 P 12 3/21 = 3/5 P 11 + P 12 2/21 + 3/21 DIT 111 PROBABILITY AND QUEUEING THEORY 108 NOTES Anna University Chennai The probability distribution of X + Y is given by X + Y P 1 0 2 P 11 = 2/21 3 P 12 + P 21 = 7/21 4 P 22 + P 31 = 8/21 5 P 32 = 5/21 Example 3 : If the joint pdf of (X,Y) is f(x, y) = 6 e -2x-3y , x ¡0, y ¡0, find the marginal density of X and conditional density of Y given X. Solution: Given f(x, y) = 6 e -2x-3y , x ¡ 0, y ¡ 0 Marginal density of X is · · · f X (x) = I f(x,y) dy = I 6 e -2x-3y dy = 6 e -2x I e -3y dy = 2e -2x ., x ¡ 0 –· 0 0 Conditional density of Y given X is given by f(Y/X) = f (x,y) = 6 e -2x-3y = 3 e -3y , y ¡ 0 f X (x) 2e -2x Example 4: the joint pdf of(X,Y) is given by f(x,y) = e -(x + y) , 0 , x,y < 8.Are X and Y are independent. Why? Solution: X and Y are independent if f (x,y) = f X (x) x f Y (y) · · · f X (x) = I f(x,y) dy = I e -(x +y) dy = e -x I e -y dy = e -x , x ¡ 0 –· 0 0 · · · f Y (y) = I f(x,y) dy = I e -(x +y) dy = e -y I e -x dy = e -y , y ¡ 0 – · 0 0 DIT 111 PROBABILITY AND QUEUEING THEORY 109 NOTES Anna University Chennai f X (x) x f Y (y) = e -x x e -y = e -(x +y) = f (x,y). Example 5: The joint probability mass function of (X,Y) is given by p(x,y) = k(2x + 3y), X = 0,1; y = 1,2,3. Find the value of k? Solution: The joint probability distribution of (X,Y) is given by X Y 1 2 3 0 3k 6k 9k 1 5k 8k 11k 2 7k 10k 13k We know that p(x, y) is a probability mass function if p ij = 1 3 2 j i p ij = 1 j =1 i = 0 i.e. the sum of all the probabilities in the table is equal to 1. i.e. 72k = 1 therefore k = 1/72. Example 6:If the joint pdf of a two –dimensional RV (X,Y) is given by f(x,y) = x 2 + (xy)/3, 0 < x < 1, 0 < y < 2 = 0, elsewhere i) Are X and Y independent? ii) Find the conditional density functions? Check whether he conditional density functions are valid ? iii) Find a)P(X > 1/2) b)P(Y < X) c) P(Y < 1/2 / X < 1/2) Solution: i)X and Y are independent if f (x,y) = f X (x) x f Y (y) · 2 f X (x) = I f(x,y) dy = I [x 2 + (xy)/3 dy] = 2x 2 + 2x/3 , 0=x=1 –· 0 · 1 f Y (y) = I f(x,y) dx = I [ x 2 + (xy)/3 ]dx = 1/3 + y/6, 0=y=2 –· 0 DIT 111 PROBABILITY AND QUEUEING THEORY 110 NOTES Anna University Chennai f X (x) x f Y (y) = 2x 2 + 2x/3 x 1/3 + y/6 | f(x,y) Therefore X and Y are not independent. ii) The conditional pdf of X given Y f(x/y) = f(x,y) = x 2 + (xy)/3 = 6x 2 + 2xy 0 = x = 1, 0 = y = 2 f Y (y) 1/3 + y/6 2 + y The conditional pdf of Y given X f(y/x) = f(x,y) = x 2 + (xy)/3 = 3x + y 0 = x = 1, 0 = y = 2 f X (x) 2x 2 + 2x/3 6x + 2 1 1 Now I f(x/y) dx = I 6x 2 + 2xy dx = 1 0 0 2 + y 2 2 Now I f(y/x) dy = I 3x + y dy = 1 0 0 6x + 2 So, the conditional density functions are valid. 2 1 iii) a) P(x > ½) = I I [x 2 + (xy)/3] dx dy = 5/6 0 ½ b) P(Y <X) The required region Y<X is shown in the following figure. From this figure we will find the limits for x &y. DIT 111 PROBABILITY AND QUEUEING THEORY 111 NOTES Anna University Chennai 1 1 P(Y <X) == I I [x 2 + (xy)/3] dx dy = 7/24 0 y c)P(Y < 1/2 / X < 1/2) = P(Y < 1/2 1 X < 1/2) P(X < 1/2) ½ ½ P(Y < 1/2 1 X < 1/2) = I I [x 2 + (xy)/3] dx dy = 5/192 0 0 2 ½ P(X < 1/2) = I I [x 2 + (xy)/3] dx dy = 1/6 0 0 Therefore P(Y < 1/2 / X < 1/2) = 5/192 = 5/32 1/6 Example 7: If f(x, y) = e -(x +y) , 0< x, y < · = 0 , otherwise is a joined pdf of RV X and Y find P(X + Y , 1) Solution: Given f(x, y) = e -(x +y) , 0< x, y < · = 0 , otherwise To find P(X + Y , 1) The required region X + Y , 1 is shown in the following figure. From this figure we will find the limits for x &y. DIT 111 PROBABILITY AND QUEUEING THEORY 112 NOTES Anna University Chennai 1 1 - y P(X + Y , 1) = I I e -(x +y) dx dy = 1 -2/e 0 · Example 8: Given the joint pdf of two R.V. (X, Y) f(x,y) = kx(x - y), 0< x < 2 , |y| < x = 0, otherwise. Evaluate the value of k? Solution: |y| < x => -x <y <x We know that · · I I f(x,y) dx dy = 1 -· -· 2 x I I kx(x - y) dx dy = 1 0 -x 8 k = 1 => k = 1/ 8 . Example 9:The two dimensional RV (X,Y) has joint density f(x,y) = 8xy, 0 < x < y < 1 = 0, otherwise i) Find P(X < ½ 1 Y < 1/4) ii) Find the marginal and conditional distributions, and iii) Are X and Y are independent? Solution: i) DIT 111 PROBABILITY AND QUEUEING THEORY 113 NOTES Anna University Chennai ¼ y P(X < ½ 1Y < 1/4)= I I 8xy dx dy = 1/256 0 0 ii) Marginal distribution of X · 1 f X (x) = I f(x,y) dy = I 8xy dy = 4x – 4x 3 , 0 < x < 1 –· x Marginal distribution of Y · y f Y (y) = I f(x,y) dy = I 8xy dx = 4 y 3 , 0 < y < 1 –· 0 Conditional density of X /Y f(x/y) = f(x,y) = 8xy = 2x , 0 < x < y < 1 f Y (y) 4 y 3 y 2 Conditional density of Y / X f(y/x) = f(x,y) = 8xy = 2y , 0 < x < y < 1 f X (x) 4 x(1 – x 2 ) (1 – x 2 ) iii)To verify X, Y are independent X and Y are independent if f (x,y) = f X (x) x f Y (y) f X (x) x f Y (y) = 4x – 4x 3 x 4 y 3 | f(x, y) Therefore X and Y are not independent. Example 10: If the joint pdf of the RV (X, Y) is given by f(x, y) = 1 . exp[-(x 2 + y 2 )/ 2σ 2 ] , -· <x, y <· , find P( X 2 + Y 2 , a 2 ) 2πσ 2 Solution: Her the entire xy-plane is the range space R and th vent space D is the interior of the circle x 2 + y 2 , a 2 . P( X 2 + Y 2 , a 2 ) = I I f(x, y)dx dy x 2 + y 2 , a 2 Transform from Cartesian system to polar system, i.e., put x = r cosθ and y = r sinθ DIT 111 PROBABILITY AND QUEUEING THEORY 114 NOTES Anna University Chennai Then dx dy = r dr dθ The domain of the integration becomes r = a 2π a Then P( X 2 + Y 2 = a 2 ) = I I 1 . exp[- r 2 / 2σ 2 ] r dr dθ 0 0 2πσ 2 2π a = 1 I exp[- r 2 / 2σ 2 ] dθ 2π 0 0 = 1 – exp(- a 2 / 2σ 2 ) Have you understood ? 1. Define joint cumulative distribution function. 2. Give the properties of cumulative distribution function. 3. Define joint probability density function. 4. Give the properties of joint probability density function 5. Define marginal probability distribution function. 6. Define marginal density function. Try yourself ! 1. Let X, Y be 2 random variables having joint pdf f(x, y) = k(x + 2y), where X, Y takes only integer values 0, 1,2. Find the marginal distribution of X & Y and conditional distribution of Y given X = 1. Solution: M D of X M D of Y X = i P i * Y = j P *j 6/27 0 3/27 1 9/27 1 9/27 2 12/27 2 15/27 f(X / Y = 1) = 4/9 f( Y / X = 1 ) = 5 /9 2. If f(x, y) = 8xy, 1< x < y < 2 9 = 0, otherwise DIT 111 PROBABILITY AND QUEUEING THEORY 115 NOTES Anna University Chennai i) find the marginal density of X and Y ii) Find he conditional density function of X, Y iii) Verify whether X & Y are independent? Solution: M D of X = 4x (4 – x 2 ) , 1 < x < 2 9 M D of Y = 4y (y 2 – 1) 9 f(x/y) = 2x . , 1< x < y < 2 (y 2 – 1) f(y/x) = 2y . , 1< x < y < 2 (4 – x 2 ) X & Y are not independent. 3. If a joint pdf f(x,y) = ( 6 – x - y) , 0 < x < 2, 2 < y < 4 8 Find i) P(X < 1 1 Y <3) ii) P( X + Y < 3) iii) P(X < 1 / y = 3) Solution: P(X < 1 1Y <3) = 3/8 ; P( X + Y < 3) = 5/24 : P(X < 1 / y = 3) = 3/5 2.5 EXPECTATION OF A FUNCTION If (X, Y) is a bivariate random variable and g(X,Y) is a function of X and Y, then E[g (x,y)] = g(x i , y j )p ij (discrete case) i j · · E[g (x,y)] = I I g(x, y) f XY (x,y)dx dy (continuous case) -· -· Note: E(X) = x i p i* ; E(Y) = y j p *j (discrete case) i j · · · · E(X) = I I x f (x,y)dxdy ; E(Y) = I I y f(x,y)dxdy (continuous case) -· -· -· -· E(X + Y) = (x i + y j )p(x i, y j ) (discrete case) i j · · E(X + Y) = I I (x + y) f XY (x,y) dx dy (continuous case) -· -· DIT 111 PROBABILITY AND QUEUEING THEORY 116 NOTES Anna University Chennai E(XY) = (x i y j )p(x i, y j ) (discrete case) i j · · E(XY) = I I (xy) f XY (x,y) dx dy (continuous case) -· -· 2.5.1 Addition Theorem of expectation: The mathematical expectation of the sum of random variables is equal to the sum of their expectations i.e., if X and Y are random Variables then E(X + Y) = E(X) + E (Y) 2.5.2 Multiplication Theorem of expectation: The mathematical expectation of the product of random variables is equal to the prod- uct of their expectations i.e., if X and Y are independent random variables then E(XY) = E(X) x E (Y) 2.5.3 Expectation of a linear combination of random variables Let X 1 , X 2 , . . . .X n be n random variables and a 1 , a 2 , a 3 , . . . a n be constants then n n E a i X i = a i E(X i ) i = 1 i =1 2.5.4 Conditional expected values If (X, Y) is a two dimensional discrete random variable with joint probability mass function P ij , then the conditional expectations of g(X,Y) are defined as follows: E{g(X,Y) / Y = Yj} = g(x i , y j ) P(X = x i / Y = y j ) = g(x i , y j ) p ij i p *j E{g(X,Y) / X = X j }= g(x i , y j ) p ij i p i* If (X, Y) is a two dimensional continuous RV with joint pdf f(x,y), then · E{g(X,Y) / Y}= I g(x, y) x f(x / y) dx -· · E{g(X,Y) / X}= I g(x, y) x f(y / x) dy -· Conditional means are defined as DIT 111 PROBABILITY AND QUEUEING THEORY 117 NOTES Anna University Chennai · µ Y/X = E( Y/X ) = I y f(y / x) dy -· · µ X/Y = E( Y/X ) = I x f(x / y) dx -· Conditional Variances are defined as · σ 2 Y/X = E{(Y - µ Y/X ) 2 } = I (y - µ Y/X ) 2 f(y / x) dy -· · σ 2 X/Y = E{(X – µ X/Y ) 2 } = I (y - µ Y/X ) 2 f(x / y) dx -· Note: If X and Y are independent RVs then E(Y/X) = E(Y) and E(X/Y) = E(X) 2. 6 COVARIANCE Let (X, Y) be a bivariate random variable. Then covariance of (X,Y) is defined as Cov(X,Y) = E[(X - X)(Y - Y)] = E[XY – XY – XY + X Y ] = E[XY] - YE[X] - X E[Y] + X Y = E[XY] - X Y - X Y + X Y = E[XY] - E[X] E[Y] Cov(X,Y) = E[XY] - E[X] E[Y] Note: If X and Y are independent then E[XY] = E[X] E[Y] and hence in this case Cov(X, Y) = 0. But the converse is not true. i.e., if Cov(X, Y) = 0 then X and Y need not be independent. (i) Cov ( aX, bY) = abCov (X,Y) Cov ( aX, bY) = E [ (aX) (bY) ] – E (aX) E (aY) = ab E (XY) – ab E(X) E(Y) = ab [ E(XY) – E (X) E(Y) ] = ab cov (X,Y) (ii) Cov ( X+a, Y+ b) = Cov (X,Y) Cov ( X+a, Y+b ) = E [ (X+a) (Y+b) – E (X+a) E(Y+b)] DIT 111 PROBABILITY AND QUEUEING THEORY 118 NOTES Anna University Chennai = E[XY+bX+aY+ab] – [E(X)+a] [E(Y)+b] = E(XY)+bE(X)+aE(Y)+ab-E(X)E(Y)-aE(Y)-bE(X)-ab = E(XY)- E(X)E(Y) = Cov (X,Y) (iii) Cov ( aX+b, cY+ b) = ac Cov (X,Y) Cov ( aX+b, cY+b ) = E [(aX+b) (cY+d)] – E (aX+b) E(cY+d)] = E[acXY+adX+bcY+bd] – [aE(X)+b] [cE(Y)+d] = acE(XY )+ adE(X)+ bcE(Y)+ bd - acE(X)E(Y) - adE(X) - bcE(Y) - bd = ac[E(XY)- E(X)E(Y)] = ac Cov (X,Y) (iv) V ( X 1 +X 2 ) = V(X 1 ) + V(X 2 ) + 2Cov(X 1 ,X 2 ) V ( X 1 +X 2 ) = E[(X 1 +X 2 ) 2 ] - E[(X 1 +X 2 ) 2 ] = E(X 1 2 +2X 1 X 2 +X 2 2 ) – [E(X 1 )+E(X 2 )] 2 = E(X 1 2 +2E(X 1 X 2 )+E(X 2 2 ) – [E(X 1 )] 2 - [E(X 2 )] 2 – 2E(X 1 )E(X 2 ) = E(X 1 2 ) – [E(X 1 )] 2 + E(X 2 2 ) – [E(X 2 )] 2 + 2[E(X 1 X 2 )- E(X 1 ) E(X 2q )] = V(X 1 ) + V(X 2 ) + 2Cov (X 1 X 2 ) (v) V ( X 1 -X 2 ) = V(X 1 ) + V(X 2 ) – 2Cov(X 1 ,X 2 ) If X 1 and X 2 are independent then V(X 1 ± X 2 ) = V(X 1 ) + V(X 2 ) Example 1: The joint pdf of (X, Y) is given by f(x, y) = 24xy, x > 0, y > 0, x + y , 1, and f(x,y) = 0 elsewhere, find the conditional mean and variance of Y given X. Solution: DIT 111 PROBABILITY AND QUEUEING THEORY 119 NOTES Anna University Chennai 1 - x f X (x) = I 24xy dxdy = 12x(1 - x) 2 , 0 < x < 1 0 f(y /x) = f (x, y) = 2y , 0< y < 1 - x f X (x) (1 - x) 2 1- x 1- x E(Y/ X = x) = I y f(y/x) dy = I 2y 2 . dy = 2/3 (1 - x) 0 0 (1 - x) 2 1- x E(Y 2 /x) = I y 2 f(y/x) dy = ½ (1 - x) 2 0 Var (Y 2 /x) = E(Y 2 /x) – [E(Y/x)] 2 = ½ (1 - x) 2 – 4/9(1 - x) 2 = 1/18(1 -x) 2 Example 2: The jont distribution of (X,Y) is given by X/ Y 1 3 9 2 1/8 1/24 1/12 4 1/4 1/4 0 6 1/12 1/24 1/12 find the Cov(X, Y) Solution: By definition cov(X,Y) = E[XY] - E[X] E[Y] E(X) = x i p(x i ) ; E(Y) = y j p(y j ) i j Marginal probability distribution of X X = i p i * = Σ P i j j =1,3, 9 2 P 21 + P 23 + P 29 = 1/8 + 1/24 + 1/12 = 1/4 4 P 41 + P 43 + P 49 = ¼ + ¼ + 0 = ½ 6 P 61 + P 63 + P 69 = 1/8 + 1/24 + 1/12 = 1/4 Marginal probability distribution of Y DIT 111 PROBABILITY AND QUEUEING THEORY 120 NOTES Anna University Chennai Y = j p * j = p i j i =2,4,6 1 P 21 + P 41 + P 61 =1/8 +1/4 + 1/8 = ½ 3 P 23 + P 43 + P 63 = 1/24 + ¼ + 1/24 = 1/3 9 P 29 + P 49 + P 69 = 1/12 +0 + 1/12 = 1/6 Now E(X) = x i p i* = 2(1/4) + 4(1/2) + 6(1/4) = ¼ E(Y) = y j p *j = 1(1/2) + 3(1/3) + 9(1/6) = 3 E(XY) = (x i y j )p(x i, y j ) = 2.1 (1/8) + 2.3 (1/24) + 2.9 (1/12) + 4.1 (1/4) + 4.3 (1/4) + 0 + 6.1 (1/8) + 6.3 (1/24) + 6.9 (1/12) = 9 Cov(X, Y) = E[XY] - E[X] E[Y] = 9 – (3.4) = -3 Example 3: Two random variables X and Y have joint pdf f XY (x,y) = xy . , 0 < x < 4, 1 < y < 5 96 0, elsewhere Find i) E(X) ii) E(Y) iii) E(XY) iv) E(2X + 3Y) v) Var(X) vi) Var(Y) and vii) Cov(X, Y) Solution: · · 5 4 E(X) = I I x f (x,y)dxdy ; = I I x (xy) dxdy = 8/3 -· -· 1 0 96 · · 5 4 E(Y) = I I y f(x,y)dx dy = I I y (xy) dxdy = 31/9 -· -· 1 0 96 · · E(XY) = I I (xy) f XY (x,y) dx dy -· -· 5 4 = I I (xy) (xy) dx dy = 248/27 1 0 96 E(2X + 3Y) = 2E(X) + 3 E(Y) = 2(8/3) + 3(31/9) = 47/3 Var (X) = E(X 2 ) – [E(X)] 2 DIT 111 PROBABILITY AND QUEUEING THEORY 121 NOTES Anna University Chennai · · 5 4 E(X 2 ) = I I x 2 f (x,y)dxdy ; = I I x 2 (xy) dxdy = 8 -· -· 1 0 96 Var (X) = E(X 2 ) – [E(X)] 2 = 8 – (8/3) 2 = 8/9 Var (Y) = E(Y 2 ) – [E(Y)] 2 · · 5 4 E(Y 2 ) = I I y 2 f (x,y)dxdy ; = I I y 2 (xy) dxdy = 13 -· -· 1 0 96 Var (Y) = E(Y 2 ) – [E(Y)] 2 = 13 – (31/9) 2 = 92/81 Cov(X, Y) = E[XY] - E[X] E[Y] = (248/27) – (8/3)(31/9) = 0 Since Cov(X, Y) = 0, X and Y are independent. Try yourself ! 1. The random variable (X, Y) has the following joint pdf f(x, y) = ½ (x + y) , 0 , x , 2, 0 , y , 2 0, otherwise a) Obtain the marginal distributions of X. b) E(X) and E(X 2 ) c) Compute Covariance (X,Y) (Solution: f(x) = x + 1; f(y) = ( 1 + y); E(X) = 14/3, E(Y) = 14/3; E(X 2 ) = 20/3, Cov(X,Y) = -60) 2. If X and Y are discrete random variables with P[X = x, Y = y] = C(x + y) for x = 0,1,2 and y = 1,2 and P[x, y] = 0, otherwise, find C and the covariance between X and Y. (Solution: C = 1/15, Cov(X, Y) = -2 / 75) 2.7 CORRELATION In probability theory and statistics, correlation, also called correlation coefficient, indicates the strength and direction of a linear relationship between two random vari- ables. In general statistic usage, correlation or co-relation refers to the departure of two variables from independence. To examine whether the two R.V.’s are inter-related, we collect n pairs of values of X and Y corresponding to n repetitions of the random experiment. Let them be (x 1 , y 1 ), (x 2 , y 2 ), ….. DIT 111 PROBABILITY AND QUEUEING THEORY 122 NOTES Anna University Chennai (x n , y n ). Then we plot the points with the co-ordinates (x 1 , y 1 ), (x 2 , y 2 ), …..(x n , y n ) on a graph paper. The simple figure consisting of the plotted points is called a scatter diagram.From the scatter diagram, we can form a fairly good, though vague idea of the relationship between X and Y. If the points are dense or closely packed, we may conclude that X and Y are correlated. On the other hand if the points are widely scat- tered throughout the graph paper. We may conclude that X and Y are either not corre- lated or poorly correlated. A number of different coefficients are used for different situations. The best known is the Pearson product-moment correlation coefficient, which is obtained by dividing the covariance of the two variables by the product of their standard deviations. Despite its name, it was first introduced by Francis Galton 2.7.1 Correlation coefficient : The correlation coefficient between two random variables X and Y is defined as r = cov (X,Y) = E(XY) – E(X)E(Y) σ X σ Y vVar(X) v Var (Y) 2.7.2 Definition : Two random variables are uncorrelated with each other, if the correlation between X and Y is equal to the product of their means. E[XY] = E(X) . E(Y) 2.7.3 Definition: The random variables are orthogonal to each other if the correlation between X and Y is equal to zero Note: Correlation coefficient is a number that varies between -1 and +1. When the correlation coefficient is 1 or -1, the random variables are perfectly correlated or have linear relationship between them. When r is 1, they are positively correlated. When r is -1, they are negatively correlated. If the correlation coefficient is 0, then the random variables are uncorrelated. If the two random variables are statistically independent then they are also uncorrelated and the covariance is zero but converse is not true. 2.7.4 KARL PEARSON FORMULAE We have the following formulae for calculating the correlation coefficient between two variables x and y, with number of data n, 1. r XY = n Σ xy - Σ x Σ y . v{n Σ x 2 – ( Σ x) 2 }v{n Σ y 2 – ( Σ y) 2 } DIT 111 PROBABILITY AND QUEUEING THEORY 123 NOTES Anna University Chennai If no assumed average is taken for x and y series. 2. r XY = Σdxdy , (Where dx = x – mean of x, dy = y – mean of y ) \Σdx 2 \Σdy 2 3. r XY = n Σd xdy - Σd x Σ dy . \{n Σd x 2 – ( Σd x) 2 }\{n Σ dy 2 – ( Σ dy) 2 } This formula is used when deviations for x and y series are taken from some assumed values. dx = x – A and dy = y – B. 4. In a bivariate frequency distribution of variables x and y correlaton coefficient is given by r XY = N Σf dxdy - Σfdx Σfdy . v{N Σfdx 2 – ( Σfdx) 2 }v{NΣfdy 2 – ( Σfdy) 2 } Note: 1.Correlation coefficient is independent of change of origin and scale. ie., If U = X- A and V = Y – B where h, k>0, then r XY = r UV h k If X and Y take considerably large values, computation of r XY will become difficult. In such problems, we may introduce change of origin and scale and compute r using the above property. 2. r XY = σx 2 + σ Y 2 – σ 2 (x - y) 2 σx σ Y 2.7. 5 Spearman’s rank correlation In statistics, rank correlation is the study of relationships between different rankings on the same set of items. It deals with measuring correspondence between two rankings, and assessing the significance of this correspondence. The Karl Pearson’s formula for calculating r is developed on the assumption that the values of the variables are exactly measurable. In some situations, it may not be possible to give precise values for the variables. In such cases we rank the observations in ascending or descending order using the numbers 1,23, . . . . n and measure the degree of relationship between the ranks instead of actual numerical values. The rank correlation coefficient when there are n ranks in each variable is given by formula ρ = 1 – 6 Σd 2 n(n 2 -1) DIT 111 PROBABILITY AND QUEUEING THEORY 124 NOTES Anna University Chennai where d = x- y is the difference between ranks of corresponding pairs x and y. n = number of observation. 2.7.5.1 Tie ranks: When the values of variables x and y are given, we can rank the values in each of the variables x and y are given, we can rank the values in each of the variables and determine the Spearman’s rank correlation coefficient. If two or more observations have the same rank we assign to them the mean rank. In this case, there is a correlation factor in the formula for ρ. The formula for V is given by, ρ = 1 - 6[ Σd 2 + Σm(m 2 -1)/12 ] n (n 2 -1) where m denotes the number of times the rank is repeated . Example 1:Two R.V X and Y take the values 1,2,3 and their probabilities are as follows: X Y 1 2 3 f(y) 1 0.1 0.1 0.1 0.3 2 0.1 0.2 0.1 0.4 3 0.1 0.1 0.1 .3 f(x) 0.3 0.4 0.3 1 Find mean, variance , E(X + Y).and correlation of X and Y (r XY ). Solution: X Y 1 2 3 p *j (= f(y)) 1 0.1 0.1 0.1 0.3 2 0.1 0.2 0.1 0.4 3 0.1 0.1 0.1 .3 p i* (= f(x)) 0.3 0.4 0.3 1 Now E(X) = x i p i* = (1 x 0.3) + (2 x 0.4) + (3 x 0.3) = 2 Therefore mean of X = 2 Now E(Y) = y j p *j = (1 x 0.3) + (2 x 0.4) + (3 x 0.3) = 2 Therefore mean of Y = 2 DIT 111 PROBABILITY AND QUEUEING THEORY 125 NOTES Anna University Chennai Now E(X 2 ) = x i 2 p i* = (1 2 x 0.3) + (2 2 x 0.4) + (3 2 x 0.3) = 4.6 E(Y 2 ) = y j 2 p *j = (1 2 x 0.3) + (2 2 x 0.4) + (3 2 x 0.3) = 4.6 Var (X) = E(X 2 ) – [E(X)] 2 = 4.6 – 4 = 0.6 Var (Y) = E(Y 2 ) – [E(Y)] 2 = 4.6 – 4 = 0.6 E(X + Y) = E(X ) + E (Y) = 2 + 2 = 4 Now Cov(X, Y) = E[XY] - E[X] E[Y] E(XY) = (x i y j )p(x i, y j ) = 1 x1 0.1 + 1 x 2 x 0.1 + 1 x 3 x 0.1 + 2 x 1 x 0.1 + 2 x 2 x 0.2 + 2 x 3 x 0.1 + 3 x 1 x 0.1 + 3 x 2 x 0.1 + 3 x 3 x 0.1 = 4 Cov(X, Y) = E[XY] - E[X] E[Y] = 4 – 4 = 0 Correlation of X and Y r XY = Cov(X, Y) σ X σ Y Cov(X, Y) = E[XY] - E[X] E[Y] σ X = \var(X) : Var (X) = E(X 2 ) – [E(X)] 2 σ Y = \var(Y) : Var (Y) = E(Y 2 ) – [E(Y)] 2 Since cov(X, Y) = 0, r XY = 0 Example 2: Suppose that 2 random variables (X, Y) has the joint pdf f(x.y) = x + y, 0 < x < 1, 0 < y < 1 0, otherwise Obtain the correlation coefficient between X and Y ? Solution : Correlation of X and Y r XY = cov(X, Y) σ X σ Y Cov(X, Y) = E[XY] - E[X] E[Y] σ X = vvar(X) : Var (X) = E(X 2 ) – [E(X)] 2 DIT 111 PROBABILITY AND QUEUEING THEORY 126 NOTES Anna University Chennai σ Y = vvar(Y) : Var (Y) = E(Y 2 ) – [E(Y)] 2 · · 1 1 E(X) = I I x f (x,y)dxdy ; = I I x (x + y) dxdy = 7/12 -· -· 0 0 · · 1 1 E(Y) = I I y f(x,y)dx dy = I I y (x + y) dxdy = 7/12 -· -· 0 0 · · 1 1 E(X 2 ) = I I x 2 f (x,y)dxdy ; = I I x 2 (x + y) dxdy = 5/12 -· -· 0 0 · · 1 1 E(Y 2 ) = I I y 2 f(x,y)dx dy = I I y 2 (x + y) dxdy = 5/12 -· -· 0 0 · · 1 1 E(XY) = I I x y f(x,y)dx dy = I I xy (x + y) dxdy = 1/3 -· -· 0 0 Var (X) = E(X 2 ) – [E(X)] 2 = 5/12 – 49/144 = 11/144 Var (Y) = E(Y 2 ) – [E(Y)] 2 = 5/12 – 49/144 = 11/144 σ X = \var(X) = \(11/144) σ Y = \var(Y) = \(11/144) r XY = Cov(X, Y) = E[XY] - E[X] E[Y] σ X σ Y σ X σ Y Cov(X, Y) = (1/3) – (7/12)(7/12) = -1/144 r XY = cov(X, Y) = - 1/144 . σ X σ Y \(11/144) \(11/144) Example 3: The random variable X has a mean value 3 and variance 2. A new random variable Y is defined as Y = 3X – 11. Check whether a) X and Y are orthogonal to each other b) X and Y are uncorrelated to each other. Solution : Given E(X) = 3 and σ 2 X = 2 Mean value of the random variable Y = E(3X – 11) = 3E(X) – 11 = 3 x 3 – 11= -2 DIT 111 PROBABILITY AND QUEUEING THEORY 127 NOTES Anna University Chennai Correlation of X and Y is E(XY) = E( X(3X - 11)) = E(3X 2 – 11X) = 3E(X 2 ) – 11 E(X) (Add and subtract 3[E(X)] 2 ) = 3E(X 2 ) – 11 E(X) + 3[E(X)] 2 ) - 3[E(X)] 2 ) = {3E(X 2 ) - 3[E(X)] 2 } + 3[E(X)] 2 – 11 E(X) = 3 Var (X) + 3[E(X)] 2 – 11 E(X) = 3 x 2 + 3(3 2 ) – 11 x 3 = 6 + 27 -33 = 0 Therefore E(XY) = 0 Since E(XY) = 0, X and Y are orthogonal. But E(XY) | E[X] E[Y] , they are correlated. Example 4: Two random variables X and Y are defined as Y = 4X + 9 . Find the correlation coefficient between X and Y. Solution: r XY = cov(X, Y) = E[XY] - E[X] E[Y] σ X σ Y σ X σ Y = E[X(4X + 9)] - E[X] E[4X + 9] = 4 E(X 2 ) + 9 E(X) – 4[E(X)] 2 – 9 E(X) σ X σ Y σ X σ Y = 4{ E(X 2 ) - [E(X)] 2 } = 4 σ X 2 = 4 σ X σ X σ Y σ X σ Y σ Y Let us find the standard deviation of Y, namely σ Y σ Y 2 = E(Y 2 ) – [E(Y)] 2 = E(4X + 9) 2 ) – [E(4X + 9)] 2 = E(16 X 2 + 72 X + 81) – [4 E(X) + 9] 2 = 16 E(X 2 ) + 72 E(X) + 81 – 16[E(X)] 2 – 72 E(X) – 81 = 16 E(X 2 ) – 16[E(X)] 2 = 16 σ X 2 Therefore r = 4 σ X = 4 σ X = 1 σ Y 4 σ X Correlation coefficient r = 1 As the correlation coefficient is 1 , they are positively perfectly correlated. Example 5 : Let the random variable X have the marginal density f(x) = 1, -½ < x < ½ and let the conditional density of Y given X, be f(y / x) = 1, x < y < x + 1, -½ < x <0 DIT 111 PROBABILITY AND QUEUEING THEORY 128 NOTES Anna University Chennai = 1, x < y< x + 1, -½ < x <0. Show that the variables X and Y are uncorrelated. Solution: ½ ½ We have E(X) = If(x) dx = I xdx = 0 -½ -½ If f(x, y) is the joint pdf of X and Y, then f(x, y) = f X (x) . f(y/x) · · E(XY) = I I xy f(x,y)dxdy -· -· 0 x+1 ½ 1-x = I I xydxdy + I I xydxdy -½ x 0 -x 0 x+1 ½ 1-x = ½ I x[y 2 ] dx + ½ I x[y 2 ] dx -½ x 0 -x = 0 Therefore Cov(X, Y) = E(XY) – E(X)E(Y) = 0 Therefore the coefficient of correlation r = 0 ¬ the variables X and Y are uncorrelated. Example 6: If X, Y and Z ate uncorrelated RVs with mean zero and SD 5,12 and 9 respectively, and if U = X + Y and V = Y+ Z, find the correlation coefficient between U and V. Solution: Since X, Y and Z are uncorrelated, Cov(X, Y) = 0 ;Cov(Y,Z) = 0; Cov(Z,X) = 0 i.e., E(XY) – E(X)E(Y) = 0, E(YZ) – E(Y)E(Z) = 0, E(ZX) – E(Z)E(X) = 0 Which means E(XY) = E(X)E(Y) E(YZ) = E(Y)E(Z) E(ZX) = E(Z)E(X) Given E(X) = 0, E(Y) = 0, E(Z) = 0 Therefore from the above equation we will have E(XY) = 0; E(YZ) = 0; E(ZX) = 0 DIT 111 PROBABILITY AND QUEUEING THEORY 129 NOTES Anna University Chennai Given SD, σ X = 5; σ Y = 12; σ Z = 9  Var(X) = 25; Var(Y) = 144; Var(Z) = 81  E(X 2 ) = 25; E(Y 2 ) = 144; E(Z 2 ) = 81 (Var(X) ) = E(X 2 ) – [E(X)] 2 but E(X) = 0 hence Var(X) = E(X 2 ) ) The correlation coefficient r UV = Cov(U, V) σ U σ V Cov(U, V) = E(UV) – E(U)E(V) E(UV) = E((X + Y) (Y +Z)) = E(XY + XZ + Y 2 + YZ) = E(X)E(Y) + E(X)E(Z) + E(Y 2 ) + E(Y). E(Z) E(U)E(V) = E(X + Y). E(Y + Z) = [E(X) + E(Y)] [E(Y) + E(Z)] = E(X)E(Y) + E(X)E(Z) + [E(Y)] 2 + E(Y). E(Z) Cov(U, V) = E(Y 2 ) – [E(Y)] 2 = Vary = 144 σ U 2 = Var (U) = Var(X +Y) = Var(X) + Var(Y) + 2Cov(X,Y) = 25 + 144 + 0 = 169  σ U = 13 σ V 2 = Var (V) = Var(Y + Z) = Var(Y) + Var(Z) + 2Cov(Y,Z) = 144 + 81 + 0 = 225  σ V = 15 Therefore r UV = 144 = 0.7385 13 x 15 Example7: let X 1 and X 2 be two independent random variables with mean 5 & 10 and SDs 2 and 3 respectively. Obtain r UV where U = 3X 1 + 4X 2 and V = 3X 1 – X 2 Solution: E(X 1 ) = 5; E(X 2 ) = 10; SD = 2 & 3 V(X 1 ) = 4 and V(X 2 ) = 9 Given U = 3X 1 + 4X 2 and V = 3X 1 – X 2 E(U) = E(3X 1 + 4X 2 ) = 3E(X 1 ) + 4E(X 2 ) = 3 x 5 + 4 x10 = 55 E(V) = E(3X 1 - X 2 ) = 3E(X 1 ) - E(X 2 ) = 3 x 5 - 10 = 5 DIT 111 PROBABILITY AND QUEUEING THEORY 130 NOTES Anna University Chennai E(UV) = E[(3X 1 + 4X 2 ) (3X 1 - X 2 )] = E(9 X 1 2 + 12X 1 X 2 – 3X 1 X 2 – 4 X 2 2 ) = 9E(X 1 2 ) + 9E(X 1 X 2 ) – 4E(X 2 2 ) E(X 1 X 2 ) = E(X 1 )E(X 2 ) (X 1 and X 2 are independent) = 5 x 10 = 50 We have V(X 1 ) = 4 E(X 1 2 ) – [E(X 1 )] 2 = 4  E(X 1 2 ) = 4 + [E(X 1 )] 2 = 4 + 25 = 29  E(X 2 2 ) = 9 + [E(X 2 )] 2 = 9 + 100 = 109 Therefore E(UV) = 9 x 29 + 9 x 50 – 4 x 109 = 275 r UV = Cov(U, V) σ U σ V Cov(U, V) = E(UV) – E(U)E(V) = 275 – 55 x 5 = 0 Therefore r UV = 0 Example 8: If (X, Y) is two dimensional RV uniformly distributed over the triangular region R bounded by y = 0 , x = 3 and y = 4/3 x. Find f X (x), f Y (y), E(X), Var(X), E(Y), Var(Y) and r XY. Solution: Since (X, Y) is uniformly distributed, f(x, y) = a constant = k Now I I f(x, y) dx dy = 1 DIT 111 PROBABILITY AND QUEUEING THEORY 131 NOTES Anna University Chennai 4 3 => I I k dx dy = 1 0 3y /4 => 6 k = 1 => k = 1/6 3 f Y (y) = I 1/6dx = (4 - y) , 0 < y < 4 3y /4 8 4x/3 f X (x) = I 1/6dy = 2x , 0 < x < 3 34 9 3 E(X) = I x f X (x) dx = I 2x 2 dx = 2 0 9 4 E(Y) = I y f Y (y) dy = I y (4 - y) dy = 4/3 0 8 3 E(X 2 ) = I x 2 f X (x) dx = I 2x 3 dx = 9/2 0 9 4 E(Y 2 ) = I y f Y (y) dy = I y 2 (4 - y) dy = 8/3 1 8 Var(X) = E(X 2 ) – [E(X)] 2 = 9/2 – 4 = 1/2 Var(Y) = E(Y 2 ) – [E(Y)] 2 = 8/3 – 16/9 = 8/9 4 3 E(XY) = I I 1/6 dx dy = 3 0 3y /4 r XY = E(XY) – E(X)E(Y) = 3 – 2 x 4/3 = ½ σx σ Y v ½ x 8/9 Example 9: f X and Y are uncorrelated random variables with variances 16 and 9 , find the correlation coefficient between X + Y and X- Y. Solution: Let U = X + Y and V = X – Y We have to find r UV = Cov(U, V) σ U σ V E(U) = E(X) + E(Y) and E(V) = E(X) – E(Y) Cov(U, V) = E(UV) – E(U)E(V) DIT 111 PROBABILITY AND QUEUEING THEORY 132 NOTES Anna University Chennai E(UV) = E((X + Y) (X -Y)) =E(X 2 – Y 2 ) = E(X 2 ) – E(Y 2 ) E(U)E(V) = E(X + Y)E(X - Y) = [E(X) + E(Y)] [E(X) – E(Y)] = [E(X)] 2 – E(X)E(Y) + E(X)E(Y) – [E(Y)] 2 = [E(X)] 2 – [E(Y)] 2 Cov(U, V) = E(X 2 ) – E(Y 2 ) - [E(X)] 2 + [E(Y)] 2 = [E(X 2 ) - [E(X)] 2 ] – [E(Y 2 ) - [E(Y)] 2 ] = Var(X) - Var(Y) = σ x 2 - σ Y 2 Var(U) = Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y) Since X and Y are uncorrelated Cov(X, Y)= 0 Var(U) = Var(X) + Var(Y) σ U 2 = σ x 2 + σ Y 2 Var(V) = Var(X - Y) = Var(X) + Var(Y) - 2Cov(X, Y) Var(V) = Var(X) + Var(Y) σ V 2 = σx 2 + σ Y 2 r UV = Cov(U, V) = σ x 2 - σ Y 2 . = σ x 2 - σ Y 2 σ U σ V \σ x 2 + σ Y 2 \σ x 2 + σ Y 2 σ x 2 + σ Y 2 = 16 - 9 = 7/25 = 0.28 16 + 9 Example10: Compute the coefficient correlation between X and Y, using the following data : X 1 3 5 7 8 10 Y 8 12 15 17 18 20 Solution: x y x 2 y 2 xy 1 8 1 64 8 3 12 9 144 36 5 15 26 225 75 7 17 49 289 119 8 18 64 324 144 10 20 100 400 200 34 90 248 1446 582 DIT 111 PROBABILITY AND QUEUEING THEORY 133 NOTES Anna University Chennai r XY = n Σ xy - Σ x Σ y . \{n Σ x 2 – ( Σ x) 2 }\{n Σ y 2 – ( Σ y) 2 } = 6 x 582 – 34 x 90 . \{6 x 248 – (34) 2 }\{6 x 1446 – (90) 2 } = 0.9879 Example11: Find the coefficient of correlation for the following data X 35 40 60 79 83 95 Y 17 28 30 32 38 49 Solution: X Y dx dy dx 2 dy 2 dxdy 35 17 -30 -15 900 225 450 40 28 -25 -4 625 16 10 60 30 -5 -2 25 4 10 79 32 14 0 196 0 0 83 38 18 6 324 36 108 95 49 30 17 900 289 510 392 194 2 2 2970 570 1178 X = 392 / 6 = 65.33 Y = 194/ 6 = 32.33 dx = x -65 dy = y-32 r = n Σ dxdy - Σdx Σ dy . \ n Σ dx 2 – (Σ dx) 2 \ n Σ dy 2 – (Σ dy) 2 = 6 x 1178 – 2 x 2 . \ 6 x 2970 – 2 2 \ 6 x 570 – 2 2 = 0.9055 Example 12 : Find the coefficient of correlation between output and cost of an automobile factory from the following data DIT 111 PROBABILITY AND QUEUEING THEORY 134 NOTES Anna University Chennai Output of cars ( in thousands) 3.5 4.2 5.6 6.5 7.0 8.2 8.8 9.0 9.7 10.0 Cost of cars ( thousand Rs.) 9.8 9.0 9.8 8.4 8.3 8.2 8.2 8.0 8.0 8.1 The correlation coefficient is unaffected by the change of origin and the scale Multiply outputs by 10 and then subtract 35. Multiply the cost ( in thousands of Rs.) by 10 and subtract 80. Solution: x y dx dy (dx) 2 (dy) 2 dxdy 0 18 -37 14 1369 196 -518 7 10 -30 6 900 36 -180 21 8 -16 4 256 16 -64 30 4 -7 0 49 0 0 35 3 -2 -1 4 1 2 47 2 10 -2 100 4 -20 53 2 16 -2 256 4 -32 55 0 18 -4 324 16 -72 62 0 25 -4 625 16 -100 65 1 28 -3 784 9 -84 375 48 5 8 4667 298 -1068 dx = x-37 dy =y - 4 r = n Σ dxdy - Σdx Σ dy . \ n Σ dx 2 – (Σ dx) 2 \ n Σ dy 2 – (Σ dy) 2 = 10 x (-1068) – 5 x 8 . \ 10 x 4667 -25 \ 10 x 298 - 64 = - 0.92 Example 13: The following table gives the bivariate frequency distribution of marks in an intelligence test obtained by 100 students according to their age : DIT 111 PROBABILITY AND QUEUEING THEORY 135 NOTES Anna University Chennai Age (x) in years Marks (y) 18 19 20 21 Total 10-20 4 2 2 0 8 20-30 5 4 6 4 19 30-40 6 8 10 11 35 40-50 4 4 6 8 22 50-60 0 2 4 4 10 60-70 0 2 3 1 6 Total 19 22 31 28 100 Calculate the coefficient of correlation between age and intelligence. Solution : Since the frequencies of various values of x and y are not equal to 1 each the formula for the computation of r XY is taken with a slight modification as given – r XY = N Σf dxdy - Σfdx Σfdy . \{N Σfdx 2 – ( Σfdx) 2 }\{NΣfdy 2 – ( Σfdy) 2 } Where dx = x-20 = u, dy = y-35 = v, 10 r XY = r UV N Σf uv - Σfu Σfv . \{N Σfu 2 – ( Σfu) 2 }\{NΣfv 2 – ( Σfv) 2 } x y Mid Points 18 19 20 21 u v 4 2 2 0 8 -16 32 20 4 2 0 -2 5 4 6 4 19 -19 19 10 2 1 0 -1 6 8 10 11 35 0 0 0 0 0 0 0 4 4 6 8 22 22 22 -4 -2 -1 0 1 0 2 4 4 10 20 40 4 -4 -2 0 2 0 2 3 1 6 18 54 -3 -6 -3 0 3 fu -38 -22 0 28 -32 fu 2 76 22 0 28 126 fuv 18 -6 0 15 27 In the above table, the figures enclosed in the circles are the values of uv 50-60 55 2 60-70 65 3 N = 100 25 167 27 19 22 31 28 40-50 45 1 Total f 20-30 25 -1 30-40 35 0 10-20 15 -2 0 1 -2 -1 Calculation of Coefficient of Correlation : 18 19 20 21 Total f fv fv 2 fuv DIT 111 PROBABILITY AND QUEUEING THEORY 136 NOTES Anna University Chennai Σfuv for the first column of the table is computed as follows = 4 x 4 + 5 x 2 + 6 x 0 + 4 x -2 + 0 x -4 + 0 x -6 = 18 Σfuv for the first row of the table is computed as follows = 4 x 4 + 2 x 2 + 2 x 0 + 0 x -2 = 20 Similarly other Σfuv values are computed. Value of ΣΣfuv obtained as the total of the entries of the last column and as that of the last row must tally. r XY = r UV = 100 x 27 – (-32) x 5 \{100 x 126 – (-32) 2 }\{100 x 167 – (5) 2 } = 0.1897 Example 14: The following are the ranks obtained by 10 students in Statistics and Mathematics Statistics 1 2 3 4 5 6 7 8 9 10 Mathematics 1 4 2 5 3 9 7 10 6 8 To what extent is the knowledge of students in the two subjects related? Solution: x y x-y = d d 2 1 1 0 0 2 4 -2 4 3 2 1 1 4 5 -1 1 5 3 2 4 6 9 -3 9 7 7 0 0 8 10 -2 9 9 6 3 9 10 8 2 4 36 The rank correlation is given by p = 1 – 6 Σd 2 n (n 2 -1) DIT 111 PROBABILITY AND QUEUEING THEORY 137 NOTES Anna University Chennai 1 – 6 X 36 = 1 – 0.219 = 0.781 10 X 99 Example 15: Ten competitors in a beauty contest are ranked by three judges in the following order First Judge 1 4 6 3 2 9 7 8 10 5 Second Judge 2 6 5 4 7 10 9 3 8 1 Third Judge 3 7 4 5 10 8 9 2 6 1 Use the method of rank correlation coefficient to determine which pair of judges have the nearest approach to common taste in beauty Solution : Let x,y,z denote the ranks by 1 st , 2 nd and 3 rd judges respectively z y z dxy(x - y) dyz(y-z) dzx(z -x) dxy 2 dyz 2 dzx 2 1 2 3 -1 -1 -2 1 1 4 4 6 7 -2 -1 -3 4 1 9 6 5 4 1 1 2 1 1 4 3 4 5 -1 -1 -2 1 1 4 2 7 10 -5 -3 -8 25 9 64 9 10 8 -1 2 1 1 4 1 7 9 9 -2 0 -2 4 0 4 8 3 2 5 2 6 25 1 36 10 8 6 2 2 4 4 4 16 5 2 1 4 0 4 16 0 16 82 22 158 ρ xy = 1 – 6 Σdxy 2 = 1 - 6 x 82 = 0.503 n (n 2 -1) 10 x 99 ρ yz = 1 – 6 Σdyz 2 = 1 - 6 x 22 = 0.867 n (n 2 -1) 10 x 99 ρ zy = 1 – 6 Σdzx 2 = 1 - 6 x 158 = 0.04 n (n 2 -1) 10 x 99 Since the rank correlation between y and z is positive highest among the three coefficients, judges y and z have the nearest approach for common taste in beauty. DIT 111 PROBABILITY AND QUEUEING THEORY 138 NOTES Anna University Chennai Example 16 : Find the rank correlation coefficient for the following data x 92 89 87 86 86 77 71 63 53 50 y 86 83 91 77 68 85 52 82 37 57 Let R 1 and R 2 denote the ranks in x and y respectively In this X series 86 is repeated twice which are in the positions 4 th and 5 th ranks. Thus common rank 4.5 ( which is the average of 4 and 5 ) is to be given for each 86. Solution: x y R 1 R 2 d = R 1 -R 2 d 2 92 86 1 2 1 1 89 83 2 4 2 4 87 91 3 1 2 4 86 77 4.5 6 1.5 2.25 86 68 4.5 7 2.5 6.25 77 85 6 3 3 9 71 52 7 9 2 4 63 82 8 5 3 9 53 37 9 10 1 1 50 57 10 8 2 4 44.50 ρ = 1 - 6[ Σd 2 + Σm(m 2 -1)/12 ] n (n 2 -1) = 1 - 6[ 44.5 + 2(2 2 -1)/12 ] = 1 - 6 x 45 = 0.727 10 x 99 990 ρ = 0.727 Example 17: Calculate the correlation coefficient for the following ages of the husbands(X) and wives(Y), using only standard deviations of X and Y: X 23 27 28 28 29 30 31 33 35 36 Y 18 20 22 27 21 29 27 29 28 29 DIT 111 PROBABILITY AND QUEUEING THEORY 139 NOTES Anna University Chennai Solution: x y dx = x-30 dy = y-24 dx 2 dy 2 d = x-y d 2 23 18 7 6 49 36 5 25 27 20 3 4 9 16 7 49 28 22 2 2 4 4 6 36 28 27 2 3 4 9 1 1 29 21 1 3 1 9 8 64 30 29 0 5 0 25 1 1 31 27 1 3 1 9 4 16 33 29 3 5 9 25 4 16 35 28 5 4 25 16 7 49 36 29 6 5 36 25 7 49 Total 0 10 138 174 50 306 σ x 2 = Σu 2 - Σ u 2 = 138/10 = 13.8 n n σ Y 2 = Σv 2 - Σ v 2 = 174/10 – (10/10) 2 = 16.4 n n σ(x – Y) 2 = σ D 2 = σ x 2 = Σd 2 - Σ d 2 = 306/10 - (50/10) 2 = 5.6 n n r XY = σ x 2 + σ Y 2 – σ 2 (x - y) = 13.8 + 16.4 – 5.6 = 0.8176 2 σ x σ Y 2 v13.8 x 16.4 How you understood ? Say true or false. 1. The variance of the sum of the random variables equals the sum of the variances if the random variables are uncorrelated. 2. Cov(X i , X j ) | Cov(X j , X i ) 3. If X 1 and X 2 are uncorrelated then X 1 and X 2 are not necessarily statistically independent. 4. If X 1 and X 2 are statistically independent then Cov (X 1 , X 2 ) = 0 (Answers: 1.true,.2.false,3.true,4.true) DIT 111 PROBABILITY AND QUEUEING THEORY 140 NOTES Anna University Chennai Short answer questions 1.Define the correlation between two random variables. 2.When are two random variables orthogonal to each other? 3. Define covariance between two random variables. 4. Define correlation coefficient. Try yourself! 1) The joint probability mass function of X and Y is given below: – x y -1 +1 0 1/8 3/8 1 2/8 2/8 Find correlation coefficient of (X,Y) (Solution : r XY = -1 . ) 3 2) Two random variables X and Y have the joint density f ( x,y ) = 2 – x - y, 0 < x < 1, 0 < y < 1. Show that cor (X,Y) = -1 . 0 otherwise 11 3) Find the coefficient of correlation between industrial production and export using the following data – Production (X) 55 56 58 59 60 60 62 Export (Y) 35 38 37 39 44 43 44 (Solution : r (X, Y) = 0.8981) 2.8 REGRESSION A regression equation allows us to express the relationship between two (or more) variables algebraically. It indicates the nature of the relationship between two (or more) variables. In particular, it indicates the extent to which you can predict some variables by knowing others, or the extent to which some are associated with others. Correlation is the study of the degree of relationship between two variables if the rela- tionship exists. Regression is the study of the relationship between the variables. If Y is the dependant variable and X is independent variable the linear relationship suggested { DIT 111 PROBABILITY AND QUEUEING THEORY 141 NOTES Anna University Chennai between the variables is called the regression equation of Y on X. This regression equation is used to estimate the value of Y corresponding to a known value of X. On the other hand, If X is the dependant variable and Y is the independent variable the linear relation expressing X in terms of Y is called the regression equation of X on Y. This is used to estimate the value of X corresponding to a known value of Y. 2.8.1 Regression lines When the random variables X and Y are linearly correlated, the points plotted on the scatter diagram, corresponding to n pairs of observed values of X and Y, will have a tendency to cluster round a straight line. The straight line is called regression line. If we treat X as the independent variable and hence assume that the values of Y depend on those of X, the regression line is called the regression line of Y on X. Thus the equation of regression line of Y on X is (Y - Y) = r σ Y (X - X ) σ X Similarly, if we assume that the values of X depend on those of the independent variable Y, the regression line of X on Y is obtained. Thus the regression equation of X on Y is (X - X) = r σ X (Y - Y ) σ Y The slope r σ Y is called the regression coefficient of Y on X and is denoted by b YX. σ X The slope r σ X is called the regression coefficient of X on Y and is denoted by b XY. σ Y Therefore the above regression lines can be written as Regression line of Y on X as (y – y ) = b YX ( x – x ) The regression equation of X on Y is (x – x ) = b XY ( y - y ) DIT 111 PROBABILITY AND QUEUEING THEORY 142 NOTES Anna University Chennai Note: b YX = r σ Y = Cov(X,Y) x σ Y = Cov(X, Y) σ X σ X σ Y σ X σ X 2 = nΣxy – ΣxΣy nΣx 2 - (Σx) 2 When deviations are taken from some assumed values, we have b YX = nΣdxdy – ΣdxΣdy nΣdx 2 – (Σdx) 2 Similarly b XY = nΣxy – ΣxΣy nΣx 2 - (Σx) 2 When deviations are taken from some assumed values, we have b XY = nΣdxdy – ΣdxΣdy nΣdy 2 – (Σdy) 2 Note: 1. The correlation coefficient is the geometric mean between the two regression coefficients b YX = r σ Y σ X b XY = r σ X σ Y b YX × b XY = r 2 r = ± \ (b YX × b XY ) r is positive if the two regression coefficients are positive. r is negative if the two regression coefficients are negative. 2.If one regression coefficient is greater than one, the other regression coefficient is less than one DIT 111 PROBABILITY AND QUEUEING THEORY 143 NOTES Anna University Chennai r 2 =1 b YX × b XY =1 b YX = 1 . b XY Therefore if one regression coefficient is greater than one, the other regression coefficient is less than one. 3.When there is perfect linear correlation between X and Y, viz., when r xy = ± 1, the two regression lines coincide. Regression curves for the means Regression curve of Y on X is y = E(Y/ X = x) Regression curve of X on Y is y = E(X/ Y = y) Example 1 : From the following data find (i) The two regression equations (ii) The coefficient between the marks in Economics and Statistics (iii) The most likely marks in statistics when marks in Economics are 30 Marks in Economics x 25 28 35 32 31 36 29 38 34 32 Marks in Statistics y 43 46 49 41 36 32 31 30 33 39 Solution: X Y dx dy dx 2 dy2 dxdy 25 43 -6 7 36 49 -42 28 46 -3 10 9 100 -30 35 49 4 13 16 169 52 32 41 1 5 1 25 5 31 36 0 0 0 0 0 36 32 5 -4 25 16 -20 29 31 -2 -5 4 25 10 38 30 7 -6 49 36 -42 34 33 3 -3 9 9 -9 32 39 1 3 1 9 3 320 380 10 20 150 438 -73 x = Σx = 320 = 32 y = Σy = 380 = 38. n 10 n 10 DIT 111 PROBABILITY AND QUEUEING THEORY 144 NOTES Anna University Chennai b YX = n Σ dxdy - Σ dx Σ dy = -73 x 10 – 10 x 20 = -0.6643 n Σ dx 2 – ( Σd x) 2 ) 10 x 150 - 100 b XY = n Σ dxdy - Σ dx Σ dy = -73 x 10 – 10 x 20 = -0.2337 n Σ dy 2 – ( Σd y) 2 ) 10 x 438 - 400 b YX = -0.6643 b XY = -0.2337 Thus the equation of regression line of Y on X is (y – y ) = b YX ( x – x ) (y - 38) = -0.6643 ( x - 32) y = -0.6643x + 59.2576 Similarly, the regression equation of X on Y is (x – x ) = b XY ( y - y ) ( x - 32) = -0.237 ( y - 38) x = -0.2337y + 40.8806 Coefficient of correlation r 2 = b XY × b YX = -0.6643 x -0.2337 = 0.1552 r = ± 0.394 Now we have to find the most likely marks in statistics (y) when marks in economics (x) are 30 y = -0.6643x + 59.2576 Put x = 30 y = -0.6643 × 30 + 59.2576 = 39 when x = 30 , y = 39 Example 2: The two lines of regression are 8x – 10y + 66 = 0 ……..(A) 40x – 18y – 214 = 0 ……..(B) DIT 111 PROBABILITY AND QUEUEING THEORY 145 NOTES Anna University Chennai The variance of x is 9. Find the (i) mean values of x and y (ii) correlation coefficient between x and y Solution: (i) since both the lines of regression passes through the mean values x and y, the point ( x, y ) must satisfy the two given regression lines. 8 x - 10 y = -66 …….(1) 40 x – 18 y = 214 …….(2) (1) X 5 ¬ 40 x – 50 y = -330…….(3) (2) X 1 ¬ 40 x – 18 y = 214 …….(4) (3) – (4) -32 y = -544 y = 17 Sub in (1) we get 8 x – 10 (17) = -66 x = 13 Hence the mean values are given by x = 13 and y = 17 (ii) Let us suppose the equation (A) is the equation of line of regression of x on y 8x = 10y – 66 i.e. x = 10 y - 66 8 i.e. b xy = 10 8 and (B) is the equation of the line of regression of y on x 18y = 40x – 214 i.e. y = 40 x – 214 18 18 i.e,. b yx = 40 18 WKT r 2 = b xy × b yx = 10 40 8 18 r 2 = 2.77 r < 1 Thus what we choose here is wrong. DIT 111 PROBABILITY AND QUEUEING THEORY 146 NOTES Anna University Chennai Hence choose equation of (A) is the the equation of line of regression of y on x and (B) equation of line of regression of x on y. (A) 10y = 8x + 66 i.e. y = 8 x + 66 10 10 i.e b yx = 8 10 (B) 40x = 18y + 214 i.e. x = 18 y + 214 40 40 i.e. b xy = 18 40 WKT r 2 = bxy × byx r 2 = 8 18 = 9 r = ± 0.6 10 40 25 Since both the regression coefficients are positive r must be positive. Hence r = 0.6 Example 3: A study of prices of rice at Chennai and Madurai gave the following data: Chennai Madurai Mean 19.5 17.75 S.D. 1.75 2.5 Also the coefficient of correlation between the two is 0.8. Estimate the most likely price of rice (i) at Chennai corresponding to the price of 18 at Madurai and (ii) at Madurai corresponding the price of 17 at Chennai. Solution: Let the price of rice at Chennai and Madurai be denoted by X and Y respectively. Then from the data x = 19.5, y = 17.75, σ X = 1.75, σ Y = 2.5 and r XY = 0.8 Thus the equation of regression line of Y on X is (y – y ) = r σ Y ( x – x ) σ X y – 17.75 = 0.8 x 2.5 ( x – 19.5) 1.75 DIT 111 PROBABILITY AND QUEUEING THEORY 147 NOTES Anna University Chennai When x = 17, y = 17.75 + 0.8 x 2.5 x (-2.5) 1.75 = 14.89 The regression equation of X on Y is (x – x ) = σ Y ( y - y ) σ X x – 19.5 = 0.8 x 1.75 ( y – 17.75) 2.5 When y = 18 x = 19.5 + 0.8 x 1.75 ( 18 – 17.75) 2.5 = 19.64 Example 4 : In a partially destroyed laboratory record of an analysis of correlation data. The data following results only are legible: Variance of X = 1. The regression equations are 3x + 2y – 26 and 6x + y = 31. What were (i) the mean values of X and Y? (ii) the standard deviation of Y? and (iii) the correlation coefficient between X and Y? Solution: (i) since the lines of regression intersect at ( x, y ), we have 3x + 2y = 26 and 6x + y = 31. Solving these equations, we get x = 4 and y = 7. (ii) which of the two equations is the regression equation of Y on X and which one is the regression equation of X on Y are not known. Let us tentatively assume that the first equation is the regression line of X on Y and second equation is the regression line of Y on X. Based on this assumption, the first equation can be re-written as x = - 2 y + 26 (1) 3 3 and the other as y = 6x + 31 (2) Then b xy = - 2 and b yx = -6 3 Thus r 2 XY = b xy × b yx = 4 DIT 111 PROBABILITY AND QUEUEING THEORY 148 NOTES Anna University Chennai Thus r XY = -2 which is wrong Hence our tentative assumption is wrong. Thus the first equation is the regression line of Y on X and re-written as y = - 3 x + 13 (3) 2 The second equation is the regression line of X on Y and re-written as x = - 1 y + 31 (4) 6 6 Hence the correct b YX = - 3 and the correct b XY = - 1 2 6 Thus r 2 XY = b xy × b yx = 1 4 Thus r XY = - 1 ( both b xy and b yx are negative ) 2 Now σ 2 Y = b yx = -3/2 = 9 σ 2 X b xy -1/6 σ 2 Y = 9 x σ 2 X = 9 σ Y = 3 Example 5: Find the angle between two lines of regression. Deduce the condition for the two lines to be i) at right angles and ii) coincident. Solution: The equations of the regression lines are (y – y ) = b YX ( x – x ) (x – x ) = b XY ( y - y ) Slope of the first line is r σ Y .= m 1 σ X Slope of the second line is r σ X .= m 2 σ Y If θ is the angle between the two lines , then tan θ = | m 1 – m 2 | 1 + m 1 m 2 DIT 111 PROBABILITY AND QUEUEING THEORY 149 NOTES Anna University Chennai = r σ Y - σ Y . σ X r σ X 1 + σ Y 2 σ X 2 = |r – 1/r| σ X σ Y σ X 2 + σ Y 2 = ( 1 – r 2 ) × σ X σ Y | r | σ X 2 + σ Y 2 The two regression lines are at right angles when θ = π/2, i.e. tanθ = · i.e, r = 0 Therefore when the linear correlation between X and Y is zero, the two lines of regression will be at right angles. The two regression lines are coincident, when θ = 0, i.e., when tanθ =0 i.e., when r = ± 1. Therefore when the correlation between X and Y is perfect, t he two regression lines will coincide. Example 6: In a bivariate population σ X = σ Y = σ and the angle between the regression line is tan -1 3, obtain the value of correlation coefficient. Solution: The angle θ between the regression lines is given by tan θ = ( 1 – r 2 ) × σ X σ Y | r | σ X 2 + σ Y 2 Using tan θ = 3 and σ X = σ Y = σ ( 1 – r 2 ) σ 2 = 3 r 2σ 2 => r 2 + 6r – 1= 0 Solving, r = -6 ± 6.325 2 Since -1 < r < , r = 0.1625 Example 7 : Calculate the co-efficient of correlation between x and y from the following table and write down the regression equation of y on x: DIT 111 PROBABILITY AND QUEUEING THEORY 150 NOTES Anna University Chennai y / x 0 - 40 - 80 - 120 - 10 – 9 4 1 0 30 - 47 19 6 0 50 - 26 18 11 0 70 - 2 3 2 2 Solution : Before doing this problem you should remember certain things. For grouped data i. ii. r XY = r UV iii. b XY | b UV b XY = b UV × (i x /i y ) , where i x is the increment in x and i y is the increment in x. iii. b YX | b VU b YX = b VU × (i y /i x ) , where i x is the increment in x and i y is the increment in x. Mean X = A + [ Σ fx × c ] , where N is the total frequency, N = Σ f N A is the mid value and C is the increment. Similarly Y= A + [ Σ f Y × c ] N In this problem we will find the regression coefficients, then coefficient of correlation and then regression equation of y on x. DIT 111 PROBABILITY AND QUEUEING THEORY 151 NOTES Anna University Chennai x y Mid Points 20 60 100 140 u v 9 4 1 14 -14 14 8 1 0 -1 47 19 6 72 0 0 0 0 0 0 26 18 11 55 55 55 -15 -1 0 1 2 3 2 2 9 18 36 8 -2 0 2 4 fu -84 0 20 4 -60 fu 2 84 0 20 8 112 fuv -21 0 14 8 1 Calculation of Coefficient of Correlation : In the above table, the figures enclosed in the circles are the values of uv -1 0 1 2 105 1 10-30 30-50 50-70 70-90 20 40 60 80 2 N =150 59 2 Total f 84 44 20 -1 0 1 Total f fv fv 2 fuv 0-40 40-80 80-120 120-160 Σfuv for the first column of the table is computed as follows = 9 x 1 + 47 x 0 + 26 x -1 + 2 x -2 = 84 Σfuv for the first row of the table is computed as follows = 9 x 1 + 4 x 0 + 1 x -1 = 8 Similarly other Σfuv values are computed. Value of ΣΣfuv obtained as the total of the entries of the last column and as that of the last row must tally. b XY = b UV × (i x /i y ) = N Σ fuv – ΣfuΣfv = 150 x 1 – -60 x 59 x ( 40 ) = 0.6015 NΣfv 2 - (Σfv) 2 150 x 105 – (59) 2 20 b YX = b VU × (i y /i x ) = N Σ fuv – ΣfuΣfv = 150 x 1 – -60 x 59 x ( 20 ) = 0.1398 NΣfu 2 - (Σfu) 2 150 x 112 – (-60) 2 40 Coefficient of correlation r = \ b YX × b XY = \ 0.6015 x 0.1398 = 0.29 Therefore r = 0.29 Now the regression equation of y on x. (y – y ) = b YX ( x – x ) (1) DIT 111 PROBABILITY AND QUEUEING THEORY 152 NOTES Anna University Chennai x = A + [ Σ fu × c ] N = 60 + [ -60 x 40] = 44 150 y = A + [ Σ fv × c ] N = 40 + [ 59 x 20] = 47.87 150 Regression equation of y on x. (y – 47.87) = 0.1398(x - 44) Y = 0.1398 x + 41.7188 How you understood ? Say true or false. 1. Correlation between variables gives the degree of relationship between them. 2. Regression between the variables gives the degree of relationship between them. 3. Regression between X and Y is same as that between Y and X. Short answer questions 1. Define regression of y on x. 2. What are regression lines. 3. Define regression coefficient. 4. Differentiate between regression and correlation. Try yourself ! 1) Given that x = 4y + 5 and y = kx + 4 are the regression lines of X on Y and Y on X respectively. Show that 0 , k , 1 . If k = 1 , find the means of X and Y and r XY . 4 16 (Solution : mean of x = 28, mean of y = 5.75, r XY = ½) 2) If the equations of the two lines of regression of y on x and x on y are respectively, 7x – 16y + 9 = 0; 5y – 4x – 3 = 0. Calculate the coefficient of correlation, x and y . (Solution: r = 0.7338, mean of x = -3/29 and mean of y = 15/29) 3) The following table gives the data on rainfall and discharge in a certain river. Obtain the line of regression of y on x. DIT 111 PROBABILITY AND QUEUEING THEORY 153 NOTES Anna University Chennai Rainfall (inches) (X): 1.53 1.78 2.60 2.95 3.42 Discharge (1000 cc) (Y): 33.5 36.3 40.0 45.8 53.5 (Solution: y = 9.7992x + 17.714) 4) The following table gives according to age X, the frequency of marks obtained Y by 100 students in an intelligence test. Measure the degree of relationship between age and intelligence test. Age Marks 16-17 17-18 18-19 19-20 30-40 20 10 2 2 40-50 4 28 6 4 50-60 0 5 11 0 60-70 0 0 2 0 70-80 0 0 0 5 (Solution: r = 0.6137) 2.9 TRANSFORMATION OF RANDOM VARIABLES 2.9.1 Two functions of two random variables If (X, Y) is a two dimensional random varialble with joint pdf f XY (x, y) and if Z = g(X, Y) and W = h(X, Y) ae two other random variables then the joint pdf of (Z, W) is given by f ZW (z,w) = |J|f XY (x, y) cx cx J = cz cw cy cy cz cw 2.9.2 One function of two random variables If a random variable Z is defined as Z = g(X, Y), where X and Y are random variables with joint pdf f(x, y).To find the pdf of Z were introduce a second random variable W = h(x, y) and obtain the joint pdf of (Z, W) by using the previous result. Let it be f Z (z)(z, w). The required pdf of Z is then obtained as the marginal pdf which is obtained by simply integrating f ZW (z, w) wiyh respect to w. DIT 111 PROBABILITY AND QUEUEING THEORY 154 NOTES Anna University Chennai · f Z (z) = If ZW (Z, W) dw -· Example1 : If X and Y are independent random variables with pdf e -x , x ¡ 0; e -y , y¡0 respectively. Find the density function of U = X , and V = X + Y. Are U and V independent? X + Y Solution: Since X and Y are independent, f XY (x, y) = e -x e -y = e –(x + y) , x,y ¡ 0 Solving equations u = x and v = x + y, we get, x + y x = uv and y = v(1 - u) cx cx J = cu cv = v u = v(1 - u) + uv = v cy cy -v 1 - u cu cv The joint pdf of (U,V) is given by f UV (u,v) = |J| f XY (x,y) = v e -(x + y) = v e -(uv + v(1 - u)) = v e -v The range space of (U, V) is obtained as follows : Since x,y ¡ 0, uv ¡ 0 and v(1- u) ¡ 0 Therefore either u ¡ 0, v ¡ 0 and 1 – u ¡ 0 i.e., 0 , u , 1 and v ¡0, or u , 0, v , 0, 1-u , 0 i.e., u , 0, u >1 which is absurd. Therefore, the range space (U, V) is given by 0 , u , 1 and v ¡0 Therefore f UV (u,v) = v e -v , 0 , u , 1 and v ¡ 0 The pdf of U is given by , · f U (u) = I f UV (u,v) dv -· DIT 111 PROBABILITY AND QUEUEING THEORY 155 NOTES Anna University Chennai · = I v e -v dv = 1 (using integration by parts I u dv = uv - Iv du -· here u = v and dv = e -v ) f U (u) = 1 i.e., U is uniformly distributed in (0, 1) The pdf of V is given by · f V (v) = I f UV (u,v) du -· · = I v e -v dv = ve -v , v = 0 -· Now f U (u) f V (v) = v e -v = f UV (u,v) Therefore U and V are independent random variables. Example 2:If X and Y are independent random variables with f X (x) = e -x U(x) and f Y (y) = 3e -3y U(y), find f Z (z), if Z = X/Y. U(X) is the unit step impulse function defined as U(X) = 1, if x ¡= 0 = 0, if x < 0 Solution: Since X and Y are independent, f XY (xy) =3e -(x + 3y) , x,y = 0 Introduce the auxiliary variable W = Y with Z = X/Y. Therefore x = zw and y = w cx cx J = cz cw cy cy cz cw w z J = 0 1 = w The joint pdf of (Z, W) is given by DIT 111 PROBABILITY AND QUEUEING THEORY 156 NOTES Anna University Chennai f ZW (z,w) = |J|f XY (x, y) = |w| ×3e -(z + 3)w ; z,w = 0 The range space is obtained as follows: Since y ¡ 0 and x ¡ 0 we have w = 0 and zw = 0 As w = 0, z = 0. The pdf is the marginal pdf, obtained by integrating f ZW (z, w) with respect to w over the range of w The pdf of z is given by f Z (z) = I 3w e -(z + 3)w dw = 3 .z ¡ 0 (Z + 3) 2 Example 3: If X and Y follow an exponential distribution with parameter 1 and are independent, find the pdf of U = X – Y Solution: Since X and Y follow an exponential distribution with parameter 1 f X (x) = e -x , x> 0 and f Y (y) = e -y , y> 0 Since X and Y are independent f XY (x, y) = e -(x + y) ; x, y > 0 Consider the auxiliary random variable V = Y along with U = X – Y x = u + v and y = v cx cx J = cu cv = 1 1 = 1 cy cy 0 1 cu cv The joint pdf of (U, V) is given by f UV (u,v) = |J| f XY (x,y) = e -(x + y) = e –(u + 2v) To find the range space of (U, V) under the transformation x = u + v and y = v DIT 111 PROBABILITY AND QUEUEING THEORY 157 NOTES Anna University Chennai Therefore the range space of (U, V) is given by v >-u , when u < 0 and v > 0 when u >0. Now the pdf of U is given by · f U (u) = I f UV (u,v) dv -· Therefore · f U (u) = I e –(u + 2v) dv when u <0 -u · = I e –(u + 2v) dv when u > 0 0 Therefore f U (u) = ½ e u , whaen u < 0 = ½ e -u , when u > 0 Example 4: If X and Y are independent random variables each following N(0, 2), find the pdf of Z = 2X + 3Y Solution: Consider the auxiliary random variable W = Y along with Z = 2X + 3Y. Therefore z = 2x + 3y and w = y x = ½ (z – 3w) and y = w cx cx J = cz cw = ½ -3/2 = ½ cy cy 0 1 cz cw Since X and Y are independent normal random variables DIT 111 PROBABILITY AND QUEUEING THEORY 158 NOTES Anna University Chennai f XY (x, y) = 1 exp[-(x 2 + y 2 )/8] , -· < x, y < ·. 8π The joint pdf of (Z, W) is given by f ZW (z, w) = |J| f XY (x, y) = ½ × 1 exp[-{(z – 3w) 2 + 4w 2 }/32] , -· < z, w < · 8π The pdf of z is · f Z (z) = 1 I exp[-{13 w 2 – 6zw + z 2 /32] dw 16π -· = 1 exp{-z 2 / 2(2\13) 2 }, -· < z < ·, which is N(0, 2\13 ) (2\13)\2vπ Example 5: If X and Y are independent random variables each normally distributed with mean zero and variance σ 2 , find the density functions of R = v(X 2 + Y 2 ) and φ = tan -1 (Y/X). Solution: Since X and Y are independent N(0, σ) f XY (x, y) = 1 exp[-(x 2 + y 2 )/ 2 σ 2 ] , -· < x, y < ·. 2π σ 2 r = v(x 2 + y 2 ) and φ = tan -1 (y/x) are the transformations from Cartesian to polar coordinates. Therefore , the inverse transformations are given by x = r cosθ and y = r sin θ J = cosθ -rsinθ = r sinθ rcosθ The joint pdf of (R, φ) is given by f Rφ (r, θ) = |r| exp[ - r 2 /2 σ 2 ] 2π σ 2 r ¡ 0, 0 , θ , 2 π The density function of R is given by 2π f R (r) = I f Rφ (r, θ) dθ = r exp(- r 2 /2 σ 2 ) 0 σ 2 r ¡ 0 DIT 111 PROBABILITY AND QUEUEING THEORY 159 NOTES Anna University Chennai The density function of φ is given by · f φ(θ) = I r exp[-r 2 /2 σ 2 ] dr 0 2π σ 2 (put t = r 2 /2 σ 2 ) · = 1 I e -t dt 2π 0 = 1/ 2π, which is a uniform distribution. Example 6: The joint pdf of a two dimensional random variable (X,Y) is given by f(x, y) = 4xy exp[-(x 2 + y 2 )]: x ¡ 0,y ¡ 0 = 0 elsewhere Find the density function U =\(X 2 + Y 2 ) Solution: Consider the auxiliary random variable V = Y along with U =\(X 2 + Y 2 ) Therefore x =\(u 2 - v 2 ) and y = v => v = 0, u = 0 and u = v  u ¡ 0 and 0 , v , u cx cx u . –v . J = cu cv = \(u 2 - v 2 ) \(u 2 - v 2 ) = u . cy cy 0 1 \(u 2 - v 2 ) cu cv The joint pdf of (U, V) is given by f UV (u,v) = |J| f XY (x,y) = = u . 4xy exp[-(x 2 + y 2 )] \(u 2 - v 2 ) = u . 4v(u 2 - v 2 ) v exp[-(u 2 )] \(u 2 - v 2 ) = 4uv exp(-u 2 ), u ¡ 0, 0 , v , u Therefore f UV (u,v) = 4uv exp(-u 2 ), u ¡ 0, 0 , v , u = 0, otherwise DIT 111 PROBABILITY AND QUEUEING THEORY 160 NOTES Anna University Chennai Now the pdf of U is given by · f U (u) = I f UV (u,v) dv -· u = I 4uv exp(-u 2 )dv = 2u 3 exp (-u 2 ) 0 Therefore pdf of U is f U (u) = 2u 3 exp (-u 2 ), u ¡ 0 = 0 , elsewhere Try yourself ! 1.The joint pdf of X and Y is given by f(x, y) = e -(x + y) , x > 0, y > 0. Find the pdf of (x +y)/2. (Solution:pdf is 4u e -2u , u ¡ 0) 2.If the pdf of a two dimensional random variable (X, Y) is given by f(x, y) = x + y, 0 , x,y , 1. Find the pdf of U = XY. (Solution: e -|u| , -· < x, y <·. ) 2 3.If X and Y are independent random variables having density functions f X (x) = 2 e -2x , x ¡ 0 and f Y (y) = 3 e -3y , y ¡ 0 0, x < 0 0, y < 0 Find the density functions of their sum U = X + Y. (solution: 6e -2u (1 – e -u ), u >0) REFERENCES: 1. T.Veerarajan, “Probability, statistics and Random Process “, Tata McGraw Hill, 2002. 2. P.Kandasamy, K. Thilagavathi and K. Gunavathi,”Probability, Random Variables and Random processors”, S. Chand, 2003. 3. S.C Gupta and V.K Kapoor,”Fundementals of Mathemetical Statistics”,Sultan Chand & Sons, 2002 DIT 111 PROBABILITY AND QUEUEING THEORY 161 NOTES Anna University Chennai UNIT 3 TESTING OF HYPOTHESES - Introduction - Tests based on normal distribution ( Large sample test ) - Student’s t test - Senedecor’s F test - χ2 test 3.1 INTRODUCTION For the purpose of determining population characteristics, instead of enumerating the entire population, the individuals in the individuals in the sample only are observed. Then the sample characteristics are utilized to estimate the characteristics of the population. For example, on examining the sample of a particular stuff we arrive at a decision of purchasing or rejecting that stuff. Sampling is quite often used in our day-to-day practical life. For example in a shop we assess the quality of the sugar, wheat or any other commodity by taking a handful of it from the bag and then decide to purchase it or not. A housewife normally tests the cooked products to find if they are properly cooked and contain the proper quantity of salt. 3.2 LEARNING OBJECTIVES The student will be able to: - List the steps of hypothesis testing. - State in your own words the type I and type II errors for a given problem. - Extract the appropriate information from a story problem to perform a complete hypothesis test. - Set up the null and alternative hypotheses correctly. - Choose the appropriate test statistic. - Choose the appropriate level of significance. DIT 111 PROBABILITY AND QUEUEING THEORY 162 NOTES Anna University Chennai - Find the critical value using a table and state the decision rule correctly. Make a statistical decision. - State the conclusion. - Perform a hypothesis test for 2 means using the appropriate formula. - Choose when to use a 2-sample t-test vs. a 2-sample z-test. - List the assumptions for a 2-sample equal (pooled) variance independent test. - Perform a 2-sample equal (pooled) variance t-test - If the problem asks for a business decision based on the hypothesis test, state the appropriate decision. - Use an F-test to perform an equality of variance hypothesis test. - Incorporate the F-test for equality of variance in the hypothesis test for 2 means. - Interpret the results of the chi-square test of independence. - Look up the critical value in the chi-square table. Generally population refers to a collection of entities such that each entity possesses an attribute called a characteristic. A statistical hypothesis, is a chain either about the value of a single population characteristics or about the values of several population characteristic.  Population A statistical population is the set of all possible measurements on data corresponding to the entire collection of units for which an inference is to be made.  Parameter and statistic You will be knowing how to find arithmetic mean, median, mode, standard deviation etc from the data contained in a sample. These are called some characterizations of a statistical distribution. These characteristics are called parameters if they are calculated for a population and are called statistics if they are calculated for a sample. For example mean of a population is called a parameter and mean of a sample is called a statistic. The values of the statistic will normally vary from one sample to another, as the values of the population members included in different samples, though drawn from the same population , may be different. These differences in the values of the statistic are said to be sampling fluctuations.  Sampling distribution. These statistics vary from sample to sample if repeated random samples of the same size are drawn from a statistical population. The probability distribution of such a statistic is called the sampling distribution. DIT 111 PROBABILITY AND QUEUEING THEORY 163 NOTES Anna University Chennai  Standard error (S.E) If a random variable X is normally distributed with mean µ and standard deviation σ then the random variable X (the mean of a simple random sample of size n) is also normally distributed with mean µ and standard deviation σ x = σ \n The standard deviation of the sampling distribution of mean referred to as the standard error of the mean and denoted by σ x = σ \n For finite population standard error of the mean is given by σ x = σ \(N – n) \n \N – 1) where N is the number of elements in the population and n is the number of elements in the sample.  Estimation and Testing of Hypothesis In sampling theory, we primarily concerned with two types of problems which are given below: a) Some characteristic or feature of the population in which we are interested may be completely unknown to us and we may like to make a guess about this characteristic entirely on the basis of a random sample drawn from the population. This type of problem is known as the problem of estimation. b) Some information regarding the characteristic or feature of the population may be available to us and we may like to know whether the information is acceptable in the light of the random sample drawn from the population and if it can be accepted, with what degree of confidence it can be accepted. This type of problem is known as the problem of testing of hypothesis. Hypothesis testing addresses the important question of how to choose among alternative propositions while controlling and minimizing the risk of wrong decisions. When we attempt to make decisions about the population on the basis of sample information , we have to make assumptions about the nature of the population involved or about the value of some parameter of the population. Such assumptions, which may or may not be true, are called statistical hypothesis. We set up a hypothesis which assumes that there is no significant difference between the sample statistic and the corresponding population parameter or between two sample statistics. Such a hypothesis of no difference is called a null hypothesis and is denoted by H 0 . DIT 111 PROBABILITY AND QUEUEING THEORY 164 NOTES Anna University Chennai A hypothesis complementary to the null hypothesis is called an alternative hypothesis and is denoted by H 1 . A procedure for deciding whether to accept or to reject a null hypothesis and hence to reject or to accept the alternative hypothesis is called the test of hypothesis.  Test of significance The difference between θ 0 and θ where θ 0 is a parameter of the population and θ is the corresponding sample statistic, which is caused due to sampling fluctuations is called insignificant difference. The difference that arises due to the reason that either the sampling procedure is not purely random or that the sample has not been drawn from the given population is known as significant difference. This procedure of testing whether the difference between θ 0 and θ is significant or not is called as the test of significance.  Critical region The critical region of a test of statistical hypothesis is that the region of the normal curve which corresponds to the rejection of null hypothesis.  Level of significance Level of significance is the probability level below which the null hypothesis is rejected. Generally, 5% and 1% level of significance are used.  Errors in hypothesis The level of significance is fixed by the investigator and as such it may be fixed at a higher level by his wrong judgment. Due to this, the region of rejection becomes larger and the probability of rejecting a null hypothesis, when it is true, becomes greater. The error committed in rejecting H 0 , when it is really true, is called Type I error. This is similar to a good product being rejected by the consumer and hence Type I error is also known as producer’s risk. The error committed in accepting H 0 , when it is false, is called Type II error. As this error is similar to that of accepting a product of inferior quality, it is also known as consumer’s risk. The probabilities of committing Type I and II errors are denoted by α & β respectively. It is to be noted that the probability of α of committing Type I error is the level of significance. DIT 111 PROBABILITY AND QUEUEING THEORY 165 NOTES Anna University Chennai  One Tailed and two tailed tests If θ 0 is a parameter of the population and θ is the corresponding sample statistic and if we set up the null hypothesis H 0 : θ = θ 0 , then the alternative hypothesis which is complementary to H 0 can be anyone of the following: i) H 1 : θ | θ 0 , i.e., θ > θ 0 or θ < θ 0 ii) H 1 : θ > θ 0 iii) H 1 : θ < θ 0 H 1 given in (i) is called a two tailed alternative hypothesis, whereas H 1 given in (ii) is called a right-tailed alternative hypothesis and H 1 given in (iii) is called a left-tailed alternative hypothesis. When H 0 is tested while H 1 is a one-tailed alternative (right or left), the test of hypothesis is called a one-tailed test. When H 0 is tested while H 1 is a two-tailed alternative (right or left), the test of hypothesis is called a two-tailed test.  Critical values or significant values The value of test statistic which separates the critical (or rejection) region and the acceptance region is called the critical value or significant value. It depends upon: i)the level of significance used and ii)the alternative hypothesis, whether it is two tails or single tailed. The critical value of the test statistic a level of significance α for a two tailed test is given by z α where z α is determined by the equation P(|Z| > z α ) = α i.e., z α is the value so that the total area of the critical region on both tails is α. Since normal probability curve is a symmetrical curve, we get P(Z > z α ) + P(Z < - z α ) = α  P(Z > z α ) + P(Z > z α ) = α  2 P(Z > z α ) = α  P(Z > z α ) = α/2  i.e., the area of each tail is α/2 . Thus z α is the value such that area to the right of z α is α/2 and to the left - z α is α/2. DIT 111 PROBABILITY AND QUEUEING THEORY 166 NOTES Anna University Chennai TWO-TAILED TEST (Level of significance α) In case of single tail alternative, the critical value z α is determined so that total area to the right of it is α and for left-tailed test the total area to the left is - z α is z α , i.e., For right tailed test : P(Z > z α ) = α For left tailed test : P(Z < - z α ) = α RIGHT-TAILED TEST (Level of significance α) DIT 111 PROBABILITY AND QUEUEING THEORY 167 NOTES Anna University Chennai LEFT-TAILED TEST (Level of significance α) Thus the significant or critical value of Z for a single value of Z for a single-tailed test at level of significance α is same as the critical value of Z for a two tailed test at level of significance ‘2 α’ The critical values z α for some standard Level of significance’s are given in the following table. Nature LOS 1%(.01) 2%(.02) 5%(.05) 10%(.1) of test Two-tailed | z α | = 2.58 | z α | = 2.33 | z α | = 1.96 | z α | = 1.645 Right-tailed zα = 2.33 z α = 2. 055 z α = 1.645 z α = 1.28 Left-tailed z α = -2.33 z α = -2.055 z α = -1.645 z α = -1.28 Procedure for testing of hypothesis 1. Null Hypothesis H 0 is defined. 2. Alternative hypothesis H 1 is also defined after a careful study of the problem and also the nature of the test(whether one Tailed or two tailed tests ) is decided. 3. LOS(Level of significance ) ‘α’ is fixed or taken from the problem if specified and z α is noted. 4. The test-statistic z = X – E(X) is computed S.E (X) 5. Comparison is made between |z| > z α , H 0 is rejected or H 1 is accepted, i.e., it is concluded that the difference between x and E(x) is significant at α LOS. DIT 111 PROBABILITY AND QUEUEING THEORY 168 NOTES Anna University Chennai  Confidence or Fiducial limits and Confidence interval Confidence interval is an interval that provides lower and upper limits for a specific population parameter is expected to lie. The two values of the statistic which determine the limits of the interval are called confidence limits. Thus confidence interval is the interval in which a population parameter is expected to lie with certain probability. For example 95% confidence interval for population mean µ is [ x - 1.96 σ , x + 1.96 σ ] \n \n 3.3 TEST BASED ON NORMAL DISTRIBUTION Tests of significance of large samples It is generally agreed that, if the size of the sample exceeds 30, it should be regarded as a large sample. The tests of significance used for large samples are different from the ones used for small samples for the reason that the following assumptions made for large sample do not hold for small samples. 1. The sampling distribution of a statistic is approximately normal, irrespective of whether the distribution of the population is normal or not. 2. Sample statistics are sufficiently close to the corresponding population parameters and hence may be used to calculate the standard error of the sampling distribution. 3.3.1 TEST 1 Test of significance of the difference between sample mean and population mean. Let X 1 , X 2 , . . . X n be the sample observations in a sample of size n, drawn from a population that is N(µ, σ) Then each X i follows N(µ, σ).Then their mean X follows a N µ, σ \n Even if the population, from which the sample is drawn, is non-normal, it is known that the above result holds good, provided n is large. Therefore the test statistic z = X - µ σ/\n As usual , if |z| , z α , the difference between the sample mean X and the population mean µ is not significant at α % LOS. DIT 111 PROBABILITY AND QUEUEING THEORY 169 NOTES Anna University Chennai Note: 1. If σ is not known, the sample S.D. ‘s’ can be used in its place, as s is nearly equal to σ when n is large. 2. 95 % confidence limits for µ are given by |X - µ | = 1.96 σ/\n i.e., [ X - 1.96 σ , X + 1.96 σ ] , if σ is known. If σ is not known, then the 95 % \n \n confidence interval is ., [ X - 1.96 s , X + 1.96 s ] \n \n 3.3.2 TEST 2 Test of significance of the difference between the means of two samples. Let X 1 and X 2 be the means of two large samples of sizes n 1 and n 2 drawn from two populations (normal or non-normal) with the same mean µ and variances σ 1 2 and σ 2 2 respectively. Then X 1 follows a N µ, σ 1 and X 2 follows a N µ, σ 2 either exactly or approximately. \n 1 \n 2 Therefore X 1 and X 2 follows a normal distribution. E(X 1 - X 2 ) = E(X 1 ) - E(X 2 ) = µ– µ = 0 V(X 1 - X 2 ) = V(X 1 ) + V(X 2 ) = σ 1 2 + σ 2 2 n 1 n 2 (since X 1 & X 2 are independent, as the samples are independent ) Thus (X 1 - X 2 ) follows a N 0, σ 1 2 + σ 2 2 n 1 n 2 Therefore the test statistic z = X 1 - X 2 σ 1 2 + σ 2 2 n 1 n 2 If |z| , z α , the difference (X 1 - X 2 ) and 0 or the difference between X 1 and X 2 is not significant at α % LOS. Note: 1. If the samples are drawn from the same population, i.e., if σ 1 = σ 2 = σ then DIT 111 PROBABILITY AND QUEUEING THEORY 170 NOTES Anna University Chennai z = X 1 - X 2 σ 1 + 1 n 1 n 2 2.If σ 1 and σ 2 are not known and σ 1 | σ 2 , σ 1 and σ 2 can be approximated by the sample S.D’s s 1 and s 2 . Therefore the test statistic z = X 1 - X 2 (a) s 1 2 + s 2 2 n 1 n 2 3.If σ 1 and σ 2 are equal and not known, then σ 1 = σ 2 = σ is approximated by σ 2 = n 1 s 1 2 + n 2 s 2 2 . Hence in such a situation , n 1 + n 2 z = X 1 - X 2 n 1 s 1 2 + n 2 s 2 2 1 + 1 n 1 + n 2 n 1 n 2 Therefore the test statistic z = X 1 - X 2 (b) s 1 2 + s 2 2 n 2 n 1 The difference in the denominators of the values of z is given in (a) and (b) may be noted. Example 1: A random sample of 200 tins of coconut oil gave an average weight of 4.95 kg with a standard deviation of 0.21 kg. Do we accept the hypothesis of net weight 5 kg per tin at 5% level ? Solution: Sample size, n = 200 Sample mean x = 4.95 kg Sample SD s = 0.21 kg Population mean µ = 5 kg The sample is a large sample and so we apply z-test H 0 : x = µ H 1 : x | µ ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 171 NOTES Anna University Chennai The test statistic z = x - µ σ/\n (when σ is not known, replace σ by s) Therefore the test statistic z = x - µ s/\n = 4 .95 - 5 = -3.37 0.21/\200 Therefore |z| = 3.37 At 1% level of significance the tabulated value of z is 2.58 H 0 is rejected at 1% level since the calculated value of |z| is greater than the table value of z. Therefore the net weight of a tin is not equal to 5 kg. Example 2:A sample of 900 items has mean 3.4 and standard deviation 2.61. Can the sample be regarded as drawn from a population with mean 3.25 at 5% level of significance? Solution: Sample size, n = 900 Sample mean x = 3.4 Sample SD s = 2.61 Population mean µ = 3.25 The sample is a large sample and so we apply z-test H 0 : x = µ H 1 : x | µ The test statistic z = x - µ σ/\n (when σ is not known, replace σ by s) Therefore the test statistic z = x - µ s/\n = 3.4 – 3.25 = 1.72 2.61/\900 Therefore |z| = 1.72 At 1% level of significance the tabulated value of z is 2.58 H 0 is accepted since the calculated value is less than the table value. Therefore it is likely that the sample belongs to the population with mean 3.4 DIT 111 PROBABILITY AND QUEUEING THEORY 172 NOTES Anna University Chennai Example 3: The mean breaking strength of the cables supplied by a manufacturer is 1800 with a SD of 100.By a new technique in the manufacturing process, it is claimed that the breaking strength of the cable has increased. In order to test this claim, a sample of 50 cables is tested and it is found that the mean breaking strength is 1850. Can we support the claim at 1% level of significance. Solution: Sample size, n = 50 Sample mean x = 1850 Population SD σ = 100 Population mean µ = 1800 The sample is a large sample and so we apply z-test H 0 : x = µ H 1 : x >µ (one-tailed test) The test statistic z = x - µ σ/\n = 1850 – 1800 = 3.54 100/\50 Therefore |z| = 1.72 At 1% level of significance the tabulated value of z is 2.33 H 0 is rejected and H 1 is accepted at 1% level since the calculated value of |z| is greater than the table value of z. i.e., is based on the sample data, we may support the claim of increase in breaking strength. Example 4: The mean value of a random sample of 60 items was found to be 145 with a SD of 40. Find the 95% confidence limits for the population mean. What size of the Sample is required to estimate the population mean within five of its actual value with 95% or more confidence, using the sample mean? Solution: Sample size, n = 60 Sample mean x = 145 Sample SD s = 40 95 % confidence limits for µ are given by | x - µ | , 1.96 σ/\n DIT 111 PROBABILITY AND QUEUEING THEORY 173 NOTES Anna University Chennai i.e., [x - 1.96 σ , x + 1.96 σ ] , if σ is known. If σ is not known, then the 95 % \n \n confidence interval is ., [ x - 1.96 s , x + 1.96 s ] \n \n i.e., [ x - 1.96 s , µ , x + 1.96 s ] \n \n i.e.,145 – 1.96 × 40 , µ , 145 + 1.96 × 40 \60 \60 i.e.,134.9 , µ ,155.1 We have to find the value of n such that P{ x - 5 = µ , x + 5} ¡ 0.95 P{ -5 , µ - x , 5} = 0.95 P{| µ - x | , 5} ¡ 0.95 P{| x - µ | , 5} ¡ 0.95 P{ | x - µ | , 5 } ¡ 0.95 σ/\n σ/\n P{ |z| , 5\n } ¡ 0.95 (1) σ where z is the standard normal variate. We know that P{ |z| , 1.96}= 0.95 Therefore the least value of n = n L that will satisfy (1) is given by 5\n L = 1.96 σ i.e., \n L = 1.96 σ = 1.96 s 5 5 n L = 1.96 × 40 2 = 245.86 5 Therefore the least size of the sample = 246 Example 5:A normal population has mean of 0.1 and SD of 2.1. Find the probability that the mean of a samples of size 900 drawn from this population will be negative. Solution: Since x follows a N µ, σ , z = | x - µ | is the standard normal variate. \n σ/\n DIT 111 PROBABILITY AND QUEUEING THEORY 174 NOTES Anna University Chennai Now P( x < 0) = P { x - 0.1 < -0.1} = P { x - 0.1 < -0.1 . } (2.1) / \900 (2.1) / \900 = P{ z < -1.43} = P{ z > 1.43} (by symmetry of the standard normal distribution) = 0.5 – P{ 0 < z < 1.43} = 0.5 - 0.4236 (from the normal table) P( x < 0) = 0.0764 Example 6: A college conducts both day and night classes intended to be identical. A sample of 100 day class students, yields examination results as mean 72.4 and SD 14.8 and a sample of 200 class students yields examination results as mean 73.9 and SD 17.9. Are the two means statistically equal at 10% level. Solution: x 1 = 72.4 x 2 = 73.9 s 1 = 14.8 s 2 = 17.91 n 1 = 100 n 2 = 200 The two given samples are large samples H 0 : µ 1 = µ 2 or x 1 = x 2 H 1 : µ 1 | µ 2 or x 1 | x 2 The test statistic z = x 1 - x 2 s 1 2 + s 2 2 n 1 n 2 = 72 .4 - 73.9 . = -0.77 \[ (14.8) 2 / 100 + (17.9) 2 /200 ] |z| = 0.77 The table value of z at 10% level = 1.645 H 0 is accepted at 10% level, since the calculated value is less than the table value. Therefore two means are statistically equal. DIT 111 PROBABILITY AND QUEUEING THEORY 175 NOTES Anna University Chennai Example 7: The sales manager of a large company conducted a sample survey in states A and B taking 400 samples in each case. The results were – State A State B Average Sales Rs.2,500.00 Rs.2,200.00 SD Rs.400.00 Rs.550.00 Test whether the average sales is the same in the two states at 1% level. Solution: x 1 = 2000 x 2 = 2200 s 1 = 400 s 2 = 550 n 1 = 400 n 2 = 400 The two given samples are large samples H 0 : µ 1 = µ 2 or x 1 = x 2 H 1 : µ 1 | µ 2 or x 1 | x 2 The test statistic z = x 1 - x 2 s 1 2 + s 2 2 n 1 n 2 = 2500 – 2200 . = 8.82 \[ (400) 2 / 400 + (550) 2 /400 ] |z| = 8.82 The table value of z at 1% level = 2.58 The calculated value of z is greater than the table value of z. Therefore H 0 is rejected at 1% level , i.e., the average sales within two states differ significantly. Example 8: In a random sample of size 500, the mean is found to be 20. In another independent sample of size 400, the mean is 15. Could the samples have been drawn from the same population with SD at 4. Solution: x 1 = 20 x 2 = 15 n 1 =500 n 2 = 400 DIT 111 PROBABILITY AND QUEUEING THEORY 176 NOTES Anna University Chennai σ = 4 The two given samples are large samples H 0 : x 1 = x 2 H 1 : x 1 | x 2 (two-tailed test) The test statistic z = x 1 - x 2 σ 1 + 1 n 1 n 2 = 20 – 15 . = 18.6 4\[ 1/ 500 + 1/400 ] |z| = 18.6 The table value of z at 1% level = 2.58 The calculated value of z is greater than the table value of z. Therefore H 0 is rejected at 1% level , i.e., the sample could not have been drawn from the same population. Example 9: Test the significance of the difference between the means of the samples, drawn from two normal populations with same SD from the following data: Sample 1 Size Mean SD 100 61 4 Sample 2 200 63 6 Solution: H 0 : µ 1 = µ 2 or x 1 = x 2 H 1 : µ 1 | µ 2 or x 1 | x 2 The population have same SD. The test statistic z = x 1 - x 2 s 1 2 + s 2 2 n 2 n 1 = 61 - 63 . = -3.02 \[ (4) 2 / 200 + (6) 2 /100 ] DIT 111 PROBABILITY AND QUEUEING THEORY 177 NOTES Anna University Chennai |z| = 3.02 The table value of z at 5% level = 1.96 The calculated value of z is greater than the table value of z. Therefore H 0 is rejected at 5% level , i.e., the two normal populations, from which the samples are drawn , may not have the same mean, though they may have the same SD. Example 10: The mean height of 50 male students who showed above average participation in college athletics was 68.2 inches with a SD of 2.5 inches, while 50 male students who showed no interest in such participation has a mean height of 67.5 inches with a SD of 2.8 inches. Test the hypothesis that male students who participated in college athletics are taller than other male students. Solution: Athletic Non-Athletic x 1 = 68.2’’ x 2 = 67.5’’ s 1 = 2.5’’ s 2 = 2.8’’ n 1 = 50 n 2 = 50 The two given samples are large samples H 0 : µ 1 = µ 2 or x 1 = x 2 H 1 : µ 1 > µ 2 or x 1 > x 2 (one-tailed test) The test statistic z = x 1 - x 2 s 1 2 + s 2 2 n 1 n 2 = 68.2 – 67.5 . = 1.32 \ [ (2.5) 2 /50 + (2.8) 2 /50 ] |z| = 1.32 The table value of z at 5% level = 1.645 The calculated value of z is less than the table value of z. Therefore H 0 is accepted and H 1 is rejected at 5% level . Therefore we cannot say that athletics are taller than non-athletics. Example 11: The average marks scored by 32 boys is 72 with a SD of 8 while that of 36 girls is 70 with a SD of 6. Test at 1% of significance whether the boys perform better than the girls. DIT 111 PROBABILITY AND QUEUEING THEORY 178 NOTES Anna University Chennai Solution: x 1 = 72 x 2 = 70 s 1 = 8 s 2 = 6 n 1 = 32 n 2 = 36 The two given samples are large samples H 0 : µ 1 = µ 2 or x 1 = x 2 H 1 : µ 1 > µ 2 or x 1 > x 2 (one-tailed test) The test statistic z = x 1 - x 2 s 1 2 + s 2 2 n 1 n 2 = 72 - 70 . = 1.15 \ [ (8) 2 /32 + (6) 2 /36 ] |z| = 1.15 The table value of z at 1% level = 2.33 The calculated value of z is less than the table value of z. Therefore H 0 is accepted and H 1 is rejected at 5% level . Therefore we cannot say that boys perform better than girls. Example 12: The heights of men in a city are normally distributed with mean 171 cm and SD 7 cm. While the corresponding value for women in the same city are 165 cm and 6 cm respectively. If a man and a women are chosen at random from this city, find the probability that the women is taller than the man. Solution: Let x 1 and x 2 denote the mean heights of men and women respectively. Then x 1 follows N( 171, 7 ) and x 2 follows a N( 165, 6). x 1 – x 2 also follows normal distribution. E(x 1 - x 2 ) = E( x 1 ) - E( x 2 ) = 171 – 165 = 6 V( x 1 - x 2 ) = V( x 1 ) + V( x 2 ) = 49 + 36 = 85 S.D of ( x 1 – x 2 ) = \85 = 9.22 DIT 111 PROBABILITY AND QUEUEING THEORY 179 NOTES Anna University Chennai S.D of ( x 1 – x 2 ) follows a N(6, 9.22) Now P( x 2 > x 1 ) = P( x 1 – x 2 < 0) = P ( x 1 – x 2 ) - 6 < - 6 . 9.22 9.22 = P {z < - 0.65}, where z is the standard normal variate. = P {z > 0.65}by symmetry. = 0.5 – P( 0 < z < 0.65) = 0.5 – 0.2422 = 0.2578. 3.3.3 TEST 3 Test of significance of the difference between sample proportion and population proportion. Let X be the number of successes in n independent Bernoulli trial in which the probability of success for each trial is a constant = p(say). Then it is known that X follows a binomial distribution with mean E(X) = np and variance V(X) = nPQ. When n is large, X follows N(np, \(nPQ)), i.e., a normal distribution with mean nP and S.D, \(nPQ), where Q = 1 – P. X follows N np , (nPQ) n n n 2 Now X is the proportion of success in the sample consisting of n trials, that is n denoted by p. Thus the sample proportion p follows N P, PQ 1 Therefore test statistic z = p - P \(PQ/n) If |z| , z α , the difference between the sample proportion p and the population mean Pis not significant at α % LOS. Note: 1. If P is not known, we assume that p is nearly equal to P and hence S.E.(p) is taken as \(pq/n) . Thus z = p - P \(pq/n) { } { } { } DIT 111 PROBABILITY AND QUEUEING THEORY 180 NOTES Anna University Chennai 2. 95% confidence limits for P are then given by | P – p | , 1.96, i.e. they are \(pq/n) [ p - 1.96v(pq/n) , p + 1.96 \(pq/n) ] 3.3.4 TEST 4 Test of significance difference between two sample proportions. Let p 1 and p 2 be the proportions of successes in two large samples of size n 1 and n 2 respectively drawn from the same population or from two population with same proportion P. Then p 1 follows N P, PQ n 1 and p 2 follows N P, PQ n 2 Therefore p 1 – p 2 , which is a linear combination of two normal variables also follows normal distribution. Now E(p 1 – p 2 ) = E(p 1 ) – E( p 2 ) = p – p = 0 V(p 1 – p 2 ) =V(p 1 ) + V(p 2 ) (since two samples are independent) = PQ 1 + 1 n 1 n 2 Therefore (p 1 – p 2 ) follows N 0, PQ 1 + 1 n 1 n 2 Therefore the test statistic z = (p 1 – p 2 ) . PQ 1 + 1 n 1 n 2 If P is not known, an unbiased estimate of P based on both samples, given by n 1 p 1 + n 2 p 2 , is used in the place of P. n 1 + n 2 If |z| , z α , the difference between the two sample proportion p 1 and p 2 is not significant at α % LOS. Example 13: While throwing 5 dice 30 times, a person obtains success 23 times ( securing a 6 was considered a success ). Can we consider the difference between the observed and the expected results as being significantly different. { } { } [ ] [ ] { } [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 181 NOTES Anna University Chennai Solution: Sample size = 5 × 30 = 150 Sample proportion , p = 23/150 Population proportion, P = 1/6 (Q = 1 – P = 5/6) H 0 : P = 1/6 H 1 : P | 1/6 The test statistic z = p - P . = 23/150 - 1/6 . = -0.438 \(PQ/n) \[1/6 × 5/6 /150] Therefore |z| = 0.438 The table value of z at 5% level = 1.96 The calculated value of z is less than the table value of z. Therefore H 0 is accepted at 5% level . Therefore the difference between the sample proportion and the population proportion is not significant. Example 14: In a certain city, 380 men out of 800 are found to be smokers. Discuss whether this information supports the view that majority of the men in this city are non smokers. Solution: Sample size = 800 Sample proportion of non-smokers, p = 420 / 800 Population proportion, P = 1/2 (Q = 1 – P = ½ ) H 0 : P = 1/2 H 1 : P > 1/6 (majority of men are non-smokers: one tail test) The test statistic z = p - P . = 420 / 800 - 1 / 2 = 1.414 \(PQ/n) \[1/2 × 1/2 /800] Therefore z = 1.414 The table value of z for one tail test at 5% level = 1.645 The calculated value of z is less than the table value of z. DIT 111 PROBABILITY AND QUEUEING THEORY 182 NOTES Anna University Chennai Therefore H 0 is accepted and H 1 is rejected at 5% level . Therefore we cannot conclude that majority are non-smokers. Example 15: Experience has shown that 20% of a manufactured product is of top quality. In one days production of 400 articles, only 50 are of top quality. Show that either the production of the day chosen was not representative sample or the hypothesis of 20% was wrong. Based on the particulars days production, find also the 95% confidence levels for the % of top quality products. Solution: Sample size = 400 Sample proportion of non-smokers, p = 50/ 400 = 1/8 Population proportion, P = (20%) = 1/5 (Q = 1 – P = 4/5 ) H 0 : P = 1/5 H 1 : P | 1/5 The test statistic z = p - P . = 1/8 – 1/5 = -3.75 \(PQ/n) \[1/5 × 4/5 /400] Therefore |z| = 3.75 The table value of z at 5% level = 1.96 The calculated value of z is greater than the table value of z. Therefore H 0 is rejected at 5% level . Therefore the production of the particular day chosen is not a representative sample. 95% confidence limits for P are then given by | P – p | , 1.96, \(pq/n) We have taken \(pq/n) in the denominator, because P is assumed to be unknown , For which we are trying to find the confidence limits and P is nearly equal to p. i.e. [ p - 1.96\(pq/n) , P , p + 1.96 \(pq/n)] i.e., 0.125 - 1.96 × \[1/8 × 7/8 /400] , P , 0.125 + 1.96 × \[1/8 × 7/8 /400] i.e., 0.093 , P , 0.157 Therefore 95% confidence limits for the percentage of top quality product are 9.3 and 15.7. Example 16: Show that for a random sample of size 100 drawn with replacement the standard error of sample proportion cannot exceed 0.05 DIT 111 PROBABILITY AND QUEUEING THEORY 183 NOTES Anna University Chennai Solution: The items of the sample are drawn one after another replacement. Therefore the proportion of success in the population, i.e., P remains a constant. We know that the sample proportion p follows N P, PQ 1 i.e., standard error of p = v(pq/n) = 1 v(pq) (n = 100) (1) 10 Now ( \P - \Q ) 2 ¡ 0 P + Q - 2vPQ ¡ 0 1 - 2\PQ ¡ 0 or \ PQ , ½ (2) using (2) in (1), we get S E of p , 1/20. that is standard error of p cannot exceed 0.05. Example 17: A cubicle die is thrown 9000 times and the throw of 3 or 4 is observed 3240 times. Show that the die cannot be regarded as an unbiased one. Solution: H 0 : the die is unbiased, i.e., P = 1/3 (= the probability of getting 3 or 4) H 1 : P | 1/3 (two tailed test) Though we may test the significance of difference between the sample and population proportions, we shall test the significance of the difference between the number X of successes in the sample and that in the population. When n is large, X follows N(np, \(nPQ)), i.e., a normal distribution with mean nP and S.D, v(nPQ), where Q = 1 – P. Therefore z = X – np \(nPQ)) = 3240 - ( 9000 × 1/3 ) = 5.37 \ [9000 × 1/3 × 2/3] |z| = 5.37 The table value of z at 5% level = 1.96 The calculated value of z is greater than the table value of z. Therefore H 0 is rejected at 5% level . Therefore the die cannot be regarded as unbiased. DIT 111 PROBABILITY AND QUEUEING THEORY 184 NOTES Anna University Chennai Example 18: A company has its head office at Calcutta and a branch at Mumbai. The personal Director wanted to know if the workers in the two places would like the introduction of a new plan of work and a survey was conducted for this purpose. Out of a sample of 500 workers at Calcutta, 62% favored the new plan. At Mumbai, out of a sample of 400 workers, 41% were against the plan. Is there any significant difference between the 2 groups in their attitude towards the new plan at 5% level. Solution: n 1 = 500 n 2 = 400 p 1 = 62/100 p 2 = 59/100 H 0 : P 1 = P 2 ( proportions in the two places are equal) H 0 : P 1 | P 2 The test statistic z = (p 1 – p 2 ) . PQ 1 + 1 n 1 n 2 If P is not known, an unbiased estimate of P based on both samples, given by n 1 p 1 + n 2 p 2 , is used in the place of P. n 1 + n 2 P = 500 × 62/100 + 400 × 59/100 = 0.607 : Q = 0.393 500 + 400 Therefore z = 0.62 - 0.59 . = 0.9146 \[0.607 × 0.393 (1/500 + 1/ 400)] The table value of z at 5% level = 1.96 The calculated value of z is less than the table value of z. Therefore H 0 is accepted at 5% level . Therefore there is no significant difference in their attitude towards the introduction of new plan. Example 19: Before increase in excise duty of tea, 400 people out of a sample of 500 persons were found to be tea drinkers. After an increase in duty 400 people were tea drinkers out of a sample of 600 people. Using the standard error of proportion state whether there is a significant decrease in the consumption of tea. [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 185 NOTES Anna University Chennai Solution: n 1 = 500 n 2 = 60 p 1 = 400/100 p 2 = 400/600 H 0 : P 1 = P 2 H 0 : P 1 > P 2 (one tail test) The test statistic z = (p 1 – p 2 ) . PQ 1 + 1 n 1 n 2 If P is not known, an unbiased estimate of P based on both samples, given by n 1 p 1 + n 2 p 2 , is used in the place of P. n 1 + n 2 P = 500 × 400/500 + 600 × 400/600 = 8/11 : Q = 3/11 500 + 600 Therefore z = 400/500 - 400/600. = 4.81 \[ 8/11 × 3/11 (1/500 + 1/ 600)] The table value of z at 1% level for a one-tail test = 2.33 The calculated value of z is greater than the table value of z. Therefore H 0 is rejected and H 1 is accepted at 1% level . Therefore there is a significant decrease in the consumption of tea after the increase in the excise duty. Example 20: 15.5 % of a random sample of 1600 under-graduates were smokers. Whereas 20% of a random sample of 900 post graduates were smokers in a state. Can we conclude that less number of under graduates are smokers than the post graduates. Solution: n 1 = 1600 n 2 = 900 p 1 = 0.155 p 2 = 0.2 H 0 : P 1 = P 2 H 0 : P 1 < P 2 (one tail test) The test statistic z = (p 1 – p 2 ) . PQ 1 + 1 n 1 n 2 [ ] [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 186 NOTES Anna University Chennai If P is not known, an unbiased estimate of P based on both samples, given by n 1 p 1 + n 2 p 2 , is used in the place of P. n 1 + n 2 P = 1600 × 0.155 + 900 × 0.2 = 0.1712 1600 + 900 Therefore z = 0.155 – 0.2 . = -2.87 \ [0.1712 × 0.8288 × (1/1600 + 1/ 900)] The table value of z at 5% level for a one-tail (left tailed ) test = -1.645 |z| > |z α | The calculated value of z is greater than the table value of z. Therefore H 0 is rejected and H 1 is accepted at 5% level . Therefore the habit of smoking is less among the undergraduates than among the postgraduates . 3.3.5 TEST 5 Test of significance of the difference between sample S.D and population S.D Let ‘s’ be the S.D of a large sample of size n drawn from a normal population with S.D σ. Then it is known that s follows a N [ σ , σ/ \(2n) ] approximately. Then the test statistic z = s - σ σ/ \(2n) As before the significance of the difference between s and σ is tested. 3.3.6 TEST 6 Test of significance of the difference between sample S.D’s of two large samples. Let s 1 and s 2 be the S.D’s of two large samples of sizes n 1 and n 2 drawn from a normal population with S.D Test of significance of the difference between sample S.D and population S.D σ. s 1 follows a N [ σ , σ/ \(2n 1 ) ] and s 2 follows a N [ σ , σ/ \(2n 2 ) ] Therefore (s 1 – s 2 ) follows N 0, σ 1 + 1 2n 1 2n 2 { } DIT 111 PROBABILITY AND QUEUEING THEORY 187 NOTES Anna University Chennai Therefore the test statistic z = s 1 – s 2 . σ 1 + 1 2n 1 2n 2 As usual, the significance of the difference between s 1 and s 2 is tested. Note: If σ is not known, it is approximated by = n 1 s 1 2 + n 2 s 2 2 . n 1 + n 2 where n 1 and n 2 are large . In this situation the test statistic z = s 1 – s 2 . n 1 s 1 2 + n 2 s 2 2 . 1 + 1 n 1 + n 2 2n 1 2n 2 z = s 1 – s 2 . s 1 2 + s 2 2 2n 2 2n 1 Example 21: A manufacturer of electric bulbs according to a certain process finds the SD of life of the lamps to be 100 hours. He wants to change the process if the new process results in a smaller variation in the life of the lamps. In adopting the new process a sample of 150 bulbs gave an SD of 95 hours. Is the manufacturer justified in changing the process. Solution: σ = 100 , n = 150 and s = 95 H 0 : s = σ H 0 : s < σ (left tailed test) Then the test statistic z = s - σ σ/ \(2n) = 95 - 100 . = -0.866 100 / v300 The table value of z at 5% level (left tailed ) = -1.645 |z| < |z α | The calculated value of z is less than the table value of z. Therefore H 0 is accepted and H 1 is rejected at 5% level . Hence the manufacturer is not justified in changing the process. { } ¯ . [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 188 NOTES Anna University Chennai Example 22: In two random samples of sizes of 150 and 250 the SD were calculated as 15.3 and 13.8. Can we conclude that the samples are drawn from the populations with the same SD. Solution: n 1 = 150 n 2 = 250 s 1 = 15.3 s 2 = 13.8 H 0 : σ 1 = σ 2 (The sample belong to the populations with same standard deviation) H 0 : σ 1 | σ 2 (The sample belong to the populations with different standard deviation) If σ is not known, the test statistic z = s 1 – s 2 . s 1 2 + s 2 2 2n 2 2n 1 = 15.3 - 13.8 . \ [ (15.3) 2 /300 – (13.8) 2 /500 ] = 1.5 / 1.0770 = 1.39 The table value of z at 5% level = 1.96 The calculated value of z is less than the table value of z. Therefore H 0 is accepted at 5% level . Hence the sample belong to the populations with the same SD. How you understood ? 1. Define sampling distribution and standard error. Obtain standard error of mean when population is large. 2. What is mean t by statistical hypothesis? What are the two types of errors of decision that arise in testing a hypotheses ? 3. Define null hypotheses and alternative hypotheses ? 4. What do you mean by critical region and acceptance region ? 5. What is the relation between critical values for a single tailed and two-tailed test. TRY YOURSELF ! 1) A sample of 400 male students is found to have a mean height 171.38 cm. Can it be reasonably regarded as sample from large population with mean height of 171.17 cm and standard deviation 3.3 cm? DIT 111 PROBABILITY AND QUEUEING THEORY 189 NOTES Anna University Chennai 2) An automatic machine fills in tea in sealed tins with mean weight of tea 1 kg and SD 1 gm. A random sample of 50 tins was examined and it was found that their mean weight was 999.50 gm. Is the machine working properly? 3) Two random samples of sizes 400 and 500 have mean 10.9 and 11.5 respectively. Can the sample be regarded as drawn from the same population with variance 25? 4) A person buys 100 electric tubes from well known makes taken at random from stocks for testing purpose. HE finds that ‘make A’ has a mean life of 1300 hours with a SD of 82 hours and ‘make B’ has mean life of 1248 hours with a SD of 93 hours. Discuss the significance of these results to test which make of electric tube should the person buy? 5) A person threw 10 dice 500 times and obtained 2560 times 4,5 or 6. Can this be attributed to fluctuation in sampling? 6) A manufacturer claimed that at least 95% of the equipment which he supplied to a factory confirmed to specifications. An examination of a sample of 200 pieces of equipment revealed that 18 were faulty. Test his claim at a significance level of 0.01, 0.05. 7) A coin was tossed 900 times and head appeared 490 times. Does the result support the hypothesis that the coin I unbiased? 8) In a sample of 600 men from a certain city, 450 men are found to be smokers. In a sample of 900 from another city 450 are found to be smokers. Do the data indicate that the two cities are significantly different with respect to prevalence of smoking habit among men? 9) A machine produced 20 defective articles in a batch of 400. After overhauling it produced 10 defectives in a batch of 300. Has the machine improved? 10) In a year there are 956 births in a town A of which 52.5% were males whereas in town A and B combined this proportion in a total of 1406 births was 0.496. Is there any significant difference in the proportion of male and female birth in the two towns? ( here n 1 = 956, n 2 = 450, p 1 = 502/956 = 0.525, p 2 = 192/450 = 0.427, P = 0.496 ) 11) The standard deviation of a sample of size 50 is 63. Could this have come from a normal population with standard deviation6? DIT 111 PROBABILITY AND QUEUEING THEORY 190 NOTES Anna University Chennai 3.4 STUDENT’S T- DISTRIBUTION Tests of significance for small samples. When the sample is small, i.e., n < 30, the sampling distributions of many statistics are not normal, even though the parent populations may be normal. Therefore the tests of significance discussed in the previous section are not suitable for small tests. Consequently we have to develop entirely different tests of significance that are applicable to small samples. Student’s t- distribution A random variable T is said to follow student’ t- distribution or simply t-distribution, if its probability density function is given by f(t) = 1 . 1 + t 2 – (v + 1) / 2 , -· < t < · \v β(v/2, 1/2 ) v v is called the number of degrees of freedom of the t-distribution. Note on degrees of freedom (d.f.): The number of independent variates which make up the statistic is known as the degrees of freedom and is usually denoted by v (the letter ‘Nu’ of the Greek alphabet.) The number of degrees of freedom, in general, is the total number of observations less the number of independent constraints imposed on the observations. t- distribution was defined by the mathematician W.S.D Gosset whose pen name is student, hence the name student’s-t distribution.  Uses of t-test This statistic is used in the following situations in tests of hypothesis. i) It is used to test whether a specific value is the population mean when the given sample is a small sample and the population S.D is not known. ii) It is also used to test the significance of difference between the means of two populations based on to small samples of sizes n 1 and n 2 when the S.D’s of the population are not known and also the samples drawn are independent. iii) It is also used to test the significance of difference between the means of paired observations. [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 191 NOTES Anna University Chennai  Properties of the sampling distribution of t. i) The probability curve of the t-distribution is similar to the standard normal curve and symmetric about t =0, bell shaped and asymptotic to the t-axis. ii) It has greater dispersion than the normal distribution. iii) It has uni-modal distribution. iv) The shape of the curve varies as the number of degrees of freedom varies. v) For sufficiently large values of n, the t-distribution tends to the standard normal distribution. vi) The mean of t-distribution is zero.  Critical Values of t and the t-table. The critical value of t at level of significance α and degrees of freedom v is given by P{|t| > t v (α)} = α for two tailed test, as in the case of normal distribution and large samples and by P{t > t v (α)} = α for the right-tailed test also, as in the case of normal distribution .The critical value of t for a single (right or left) tailed test at LOS ‘α’ corresponding to v degrees of freedom is the same as that for a two-tailed test at LOS ‘2 α’ corresponding to the same degrees of freedom. Critical values t v (α) of the t-distribution for two-tailed tests corresponding to a few important levels of significance and a range of values of v have been published by Prof. R.A.Fisher in the form of a table, called the t-table. DIT 111 PROBABILITY AND QUEUEING THEORY 192 NOTES Anna University Chennai 3.4.1 TEST 1 Test of significance of the difference between sample mean and population mean. If x is the mean of a sample of size n, and s is the sample standard deviation the test statistic is given by t = x - µ . s / \(n -1) This t-statistic follows a t-distribution with number of degrees of freedom v = n – 1. Sometimes t = x - µ . S / \n n where S 2 = Σ ( x r - x ) 2 and is called student’s-t. r = 1 n -1 We shall use only t = x - µ ., where s is the sample S.D. s / \(n -1) We get the value of t v (α) for the LOS α and v = n – 1.from the table. If the calculated value of t satisfies |t| > t v (α), the null hypothesis H 0 is accepted at LOS ‘α’ otherwise, H 0 is rejected at LOS α . Note: 95% confidence interval of m is given by = x - µ . , t 0.05 , since s / \(n -1) P x - µ . , t 0.05 = 0.95 s / \(n -1) i.e., by x - t 0.05 s , m , x + t 0.05 s where t 0.05 is the 5 % critical value of \(n -1 \(n -1 for n – 1 degrees of freedom for a two tailed test. 3.4.2 TEST 2 Test of significance of the difference between means of two small samples drawn from the same normal population. The test statistic is given by t = x 1 – x 2 . S 1 + 1 n 1 n 2 { } DIT 111 PROBABILITY AND QUEUEING THEORY 193 NOTES Anna University Chennai where S 2 = n 1 s 1 2 + n 2 s 2 2 n 1 + n 2 – 2 n 1 + n 2 – 2 is the number of degrees of freedom of the statistic. Note: If n 1 = n 2 = n and if the pairs of values of X 1 and X 2 are associated in some way or correlated we shall assume that H 0 : d (= x - y ) = 0 and test the significance of the difference between d and 0, using the test statistic t = d . with v = n- 1, s / \(n -1) where d i = x i - y i ( i = 1,2, …n ), d = x – y: and n s = S.D of d’s = 1 Σ (d i – d) 2 n i = 1 Example 1: A sample of 10 house owners is drawn and the following values of their incomes are obtained. Mean Rs.6,000.00 ; SD Rs. 650.00. test the hypothesis that the average income of house owners of the town is Rs.5,500.00. Solution: n = 10 s = 650 x = 6,000 µ 0 = 5,500 since the sample size n = 10 < 30, the sample is a small sample. Therefore we have to apply t-test for testing the mean. H 0 : x = µ (ie the average income of the house owners of the town is Rs.5,500) H 1 : x | µ If x is the mean of a sample of size n, and s is the sample standard deviation the test statistic is given by t = x - µ . s / \(n -1) This t-statistic follows a t-distribution with number of degrees of freedom v = n – 1. t = 6,000 – 5,500 = 2.31 650 / \9 Number of degrees of freedom = n – 1 = 9 The table value of t for 9 degrees of freedom at 5% level = 2.262 H 0 is rejected since the calculated value of t > the table value of t. Hence the average income of house owners in that tow is not Rs.5,500/- DIT 111 PROBABILITY AND QUEUEING THEORY 194 NOTES Anna University Chennai Example 2: A machine is designed to produce insulating washers for electrical devices of average thickness of 0.025 cm. A random sample of 10 washers was found to have an average thickness of 0.024 cm with a standard deviation of 0.002 cm. Test the significance of deviation. Solution: n = 10 s = 0.002 x = 0.024 cm µ = 0.025cm since the sample size n = 10 < 30, the sample is a small sample. Therefore we have to apply t-test for testing the mean. H 0 : x = µ H 1 : x | µ If x is the mean of a sample of size n, and s is the sample standard deviation the test statistic is given by t = x - µ . s / \(n -1) This t-statistic follows a t-distribution with number of degrees of freedom v = n – 1. t = 0.024 – 0.025 = -1.5 0.002 / \9 | t | = 1.5 Number of degrees of freedom = n – 1 = 9 The table value of t for 9 degrees of freedom at 5% level = 2.262 H 0 is accepted since the calculated value of |t| < the table value of t. Hence deviation is not significant. Example 3: The mean lifetime of 25 bulbs is found as 1550 hours with a SD of 120 hours. The company manufacturing the bulbs claims that the average life of their bulbs is 1600 hours. Is this claim acceptable at 5% level of significance? Solution: n = 25 s = 120 x = 1550 µ = 1600 since the sample size n = 10 < 30, the sample is a small sample. Therefore we have to apply t-test for testing the mean. DIT 111 PROBABILITY AND QUEUEING THEORY 195 NOTES Anna University Chennai H 0 : x = µ H 1 : x < µ (left tailed test) If x is the mean of a sample of size n, and s is the sample standard deviation the test statistic is given by t = x - µ . s / \(n -1) This t-statistic follows a t-distribution with number of degrees of freedom v = n – 1. t = 1550 – 1600 = -2.04 120 / \24 | t | = 2.04 Number of degrees of freedom = n – 1 = 24 The table value of t for 24 degrees of freedom at 5% level for one-tailed test = = The table value of t for 24 degrees of freedom at 10% level for two-tailed test = 1.71 H 0 is rejected and H 1 is accepted since the calculated value of |t| > the table value of t. Therefore the claim of company cannot be accepted at 5% LOS. Example 4: A filling machine is expected to fill 5 kg of powder into bags. A sample of 10 bags gave the weights 4.7, 4.9, 5.0, 5.1, 5.4, 5.2, 4.6, 5.1, 4.6 and 4.7. test whether the machine is working properly. Solution: n = 10 µ = 5 kg Let us calculate x and s from the sample data x x 2 4.7 22.09 4.9 24.01 5.0 25.00 5.1 26.01 5.4 29.16 5.2 27.04 4.6 21.16 5.1 26.01 4.6 21.16 4.7 22.09 49.3 243.73 DIT 111 PROBABILITY AND QUEUEING THEORY 196 NOTES Anna University Chennai x = Σ x / n = 49.3/ 10 = 4.93 s 2 = Σ x 2 - Σ x 2 n n s 2 = 243.73/10 – (4.93) 2 s = \0.073 = 0.27 since the sample size n = 10 < 30, the sample is a small sample. Therefore we have to apply t-test for testing the mean. H 0 : x = µ H 1 : x | µ If x is the mean of a sample of size n, and s is the sample standard deviation the test statistic is given by t = x - µ . s / \(n -1) This t-statistic follows a t-distribution with number of degrees of freedom v = n – 1. t = 4.93 - 5 = -0.78 0.27 / \9 | t | = 0.78 Number of degrees of freedom = n – 1 = 9 The table value of t for 9 degrees of freedom at 5% level for one-tailed test = 2.262 H 0 is accepted since the calculated value of |t| < the table value of t. Hence the machine is working properly. Example 5: The heights of 10 males of a given locality are found to be 175, 168, 155, 170, 152, 170, 175, 160 and 165 cms. Based on this sample of 10 items, test the hypothesis that the mean height of males is 170 cms. Also find the 95% confidence levels for the height of the males in that locality. Solution: n = 10 µ = 170 ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 197 NOTES Anna University Chennai Let us calculate x and s from the sample data x d d 2 175 10 100 168 3 9 155 -10 100 170 5 25 152 -13 169 170 5 25 175 10 100 160 -5 25 160 -5 25 165 0 0 1650 0 578 x = Σ x / n = 1650/ 10 = 165 s 2 = Σ d 2 - Σ d 2 n n s 2 = 578/10 – 0 s = \57.8 = 7.6 since the sample size n = 10 < 30, the sample is a small sample. Therefore we have to apply t-test for testing the mean. H 0 : x = µ H 1 : x | µ If x is the mean of a sample of size n, and s is the sample standard deviation the test statistic is given by t = x - µ . s / \(n -1) This t-statistic follows a t-distribution with number of degrees of freedom v = n – 1. t = 165 - 170 = -1.97 7.6/ \9 | t | = 1.97 Number of degrees of freedom = n – 1 = 9 The table value of t for 9 degrees of freedom at 5% level for = 2.26 ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 198 NOTES Anna University Chennai H 0 is accepted since the calculated value of |t| < the table value of t. This means the mean height of males can be regarded as 170 cm. 95% confidence interval of m is given by = x - µ . , t 0.05 , s / \(n -1) i.e., by [ x - t 0.05 s , m , x + t 0.05 s ] where t 0.05 is the 5 % critical value of t \n -1 \n -1 for n – 1 degrees of freedom for a two tailed test. 165 – 2.26 7.6 , m , 165 + 2.26 7.6 \9 \9 159.3 , m , 170.7 i.e., the heights of males in the locality are likely to lie within 159.3 cm and 170.7 cm. Example 6 : A certain injection administered to each 12 patients resulted in the following increases of blood pressure : 5, 2, 8, -1, 3, 0, 6, -2, 1, 5, 0, 4. Can it be concluded that the injection will be, in general, accompanied by an increase in B.P?. Solution: n = 12 µ = 1600 Let us calculate x and s from the sample data x x 2 5 25 2 4 8 64 -1 1 3 9 0 0 6 36 -2 4 1 1 5 25 0 0 4 16 31 185 DIT 111 PROBABILITY AND QUEUEING THEORY 199 NOTES Anna University Chennai x = Σ x / n = 31 / 12 = 2.58 s 2 = Σ x 2 - Σ x 2 n n s 2 = 185/12 – (2.58) 2 s = \2.96 since the sample size n = 10 < 30, the sample is a small sample. Therefore we have to apply t-test for testing the mean. H 0 : x = µ (where µ = 0 i.e., the injection will not result in increase in B.P) H 1 : x > µ (Right tailed test) If x is the mean of a sample of size n, and s is the sample standard deviation the test statistic is given by t = x - µ . s / \(n -1) This t-statistic follows a t-distribution with number of degrees of freedom v = n – 1. t = 2.58 - 0 = 2.89 2.96 / \11 | t | = 2.89 Number of degrees of freedom = n – 1 = 11 The table value of t for 11 degrees of freedom at 5% level for one-tailed test = = The table value of t for 11 degrees of freedom at 10% level for two-tailed test = 1.80 H 0 is accepted and H 1 is accepted since the calculated value of |t| < the table value of t. i.e., we may conclude that the injection is accompanied by an increase in B.P. Example 7: Two samples of 6 and 5 items respectively gave the following data : Mean of the 1 st sample = 40 SD of the 1 st sample = 8 Mean of the 2 st sample = 50 SD of the 2 st sample = 10 Is the difference between the means significant? The value of t for 9df at 5% level is 2.26. Solution: The two given samples are small samples. Let us apply t-test for testing the mean. ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 200 NOTES Anna University Chennai H 0 :µ 1 = µ 2 (The means of the two population are equal) H 1 : µ 1 ? µ 2 (The means of the two population are not equal) n 1 = 6 n 2 = 5 x 1 = 40 x 2 = 50 s 1 = 8 s 2 = 10 The test statistic is given by t = x 1 – x 2 . S 1 + 1 n 1 n 2 where S 2 = n 1 s 1 2 + n 2 s 2 2 n 1 + n 2 – 2 n 1 + n 2 – 2 is the number of degrees of freedom of the statistic. S 2 = 6 × 64 + 5 ×100 = 98.22 6 + 5 – 2 S = 9.91 Therefore t = 40 - 50 . = -1.65 9.91\( 1/6 + 1/5) |t| = 1.65 number of degrees of freedom, ndf = 6 + 5 – 2 = 9 The table value of t for 9 df at 5% level = 2.262 The calculated value of t < the table value of t. H 0 is accepted at 5% level. Hence there is no significant difference between the means of the population. Example 8: Below are given the gains in weights ( lbs ) of cows fed on two diets X and Y. Gain in weight ( in lbs ) Diet X 25 32 30 32 24 14 32 Diet Y 24 34 22 30 42 31 40 30 32 35 Test at 5% level, whether the two diets differ as regards their effect on mean increase in weight ( table value of t for 15df at 5% is 2.131 ) Solution: The two given samples are small samples. Let us apply t-test for testing the mean. DIT 111 PROBABILITY AND QUEUEING THEORY 201 NOTES Anna University Chennai H 0 :µ 1 = µ 2 (The means of the two population are equal) H 1 : µ 1 | µ 2 (The means of the two population are not equal) Let us calculate the mean and S.D of the two samples – x d 1 (x-27) d 1 2 y d 2 (y-32) d 2 2 25 -2 4 24 -8 64 32 5 25 34 2 4 30 3 9 22 -10 100 32 5 25 30 -2 4 24 3 9 42 10 100 14 -5 169 31 -1 1 32 25 40 8 64 30 -2 4 32 0 0 35 3 9 189 266 320 350 n 1 = 7 n 2 = 10 x = Σ x / n 1 = 189 / 7= 27 y = Σ y / n 2 = 320 / 10 = 32 Let d 1 = x – 27 d 2 = y – 32 s 1 2 = Σ d 1 2 - Σ d 1 2 n 1 n 1 s 1 2 = 266/7 – 0 = 38 s 2 2 = Σ d 2 2 - Σ d 2 2 n 2 n 2 s 2 2 = 350/10 = 35 The test statistic is given by t = x 1 – x 2 . S 1 + 1 n 1 n 2 where S 2 = n 1 s 1 2 + n 2 s 2 2 n 1 + n 2 – 2 n 1 + n 2 – 2 is the number of degrees of freedom of the statistic. ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 202 NOTES Anna University Chennai S 2 = 7 × 38 + 10 ×35 = 41.07 7 + 10 – 2 S = 6.41 Therefore t = 27 - 32 . = -1.59 6.41 \( 1/7 + 1/10) |t| = 1.59 number of degrees of freedom, ndf = 7 + 10– 2 = 15 The table value of t for 15 df at 5% level = 2.131 The calculated value of t < the table value of t. H 0 is accepted at 5% level. Hence there is no significant difference between the mean increase in the weight due to two diets. Example 9 : A group of 5 patients treated with medicine A weigh 42, 39, 48, 60 and 41 kgs; A second group of 7 patients from the same hospital treated with medicine B weigh 38, 42, 56, 64, 68, 69 and 62 kgs. Do you agree with the claim that the medicine B increases weight significantly. ( the value of t at 5% significance for 10df is 2.228 ) Solution: The two given samples are small samples. Let us apply t-test for testing the mean. H 0 :µ 1 = µ 2 H 1 : µ 1 < µ 2 (Medicine B increases significantly) (one tailed test) n 1 = 5 n 2 = 7 x d 1 (x-46) d 1 2 y d 2 (y-57) d 2 2 42 -4 16 38 -19 361 39 -7 49 42 -15 225 48 2 4 56 -1 1 60 14 196 64 7 49 41 -5 25 68 11 121 69 12 144 230 0 290 399 0 926 x = Σ x / n 1 = 230 / 5 = 46 y = Σ y / n 2 = 399 / 7 = 57 DIT 111 PROBABILITY AND QUEUEING THEORY 203 NOTES Anna University Chennai Let d 1 = x –46 d 2 = y – 57 s 1 2 = Σ d 1 2 - Σ d 1 2 n 1 n 1 s 1 2 = 290 / 5 – 0 = 290 / 5 s 2 2 = Σ d 2 2 - Σ d 2 2 n 2 n 2 s 2 2 = 926 / 7 – 0 = 926 / 7 The test statistic is given by t = x 1 – x 2 . S 1 + 1 n 1 n 2 where S 2 = n 1 s 1 2 + n 2 s 2 2 n 1 + n 2 – 2 n 1 + n 2 – 2 is the number of degrees of freedom of the statistic. S 2 = 5 × 58 + 10 × 926 / 7 = 121.6 5 + 7 – 2 S = 11.03 Therefore t = 46 - 57 . = -1.7 11.03v( 1/5 + 1/7) |t| = 1.7 number of degrees of freedom, ndf = 5 + 7 – 2 = 10 The table value of t for 10 df at 5% level for one tailed test = The table value of t for 10 df at 10% level for two tailed test = 1.812 The calculated value of t < the table value of t. H 0 is accepted at 5% level. Therefore medicine A and B do not differ significantly w.r.t increase in weights. Example 10 : The marks obtained by a group of 9 regular course students and another group of 11 part time course students in a test are given below – Regular 56 62 63 54 60 51 67 69 58 Part time 62 70 71 62 60 56 75 64 72 68 66 ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 204 NOTES Anna University Chennai Examine whether the marks obtained by regular students and part time students differ significantly at 5% level of significance and 1% level of significance. Solution: The two given samples are small samples. Let us apply t-test for testing the mean. H 0 :µ 1 = µ 2 H 1 : µ 1 | µ 2 Let us calculate the mean and variance of the two samples. x d 1 d 1 2 y d 2 (y-57) d 2 2 56 -4 16 62 -4 16 62 2 4 70 4 16 63 3 9 71 5 25 54 -6 36 62 -4 16 60 0 0 60 -6 36 51 -9 81 56 -10 100 67 7 49 75 9 81 69 9 81 64 -2 4 58 -2 4 72 6 36 68 2 4 66 0 0 540 0 280 726 0 334 n 1 = 9 n 2 = 11 x = Σ x / n 1 = 230 / 5 = 46 y = Σ y / n 2 = 399 / 7 = 57 Let d 1 = x –46 d 2 = y – 57 s 1 2 = Σ d 1 2 - Σ d 1 2 n 1 n 1 s 1 2 = 280 / 9 – 0 = 280 / 9 ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 205 NOTES Anna University Chennai s 2 2 = Σ d 2 2 - Σ d 2 2 n 2 n 2 s 2 2 = 334 /11 – 0 = 334 /11 The test statistic is given by t = x 1 – x 2 . S 1 + 1 n 1 n 2 where S 2 = n 1 s 1 2 + n 2 s 2 2 n 1 + n 2 – 2 n 1 + n 2 – 2 is the number of degrees of freedom of the statistic. S 2 = 9× 280 / 9 + 11 × 334 /11 = 34.11 9 + 11 – 2 S = 5.84 Therefore t = 60 - 66 . = -2.28 5.84v( 1/9 + 1/11) |t| = 2.28 number of degrees of freedom, ndf = 9 + 11 – 2 = 18 The table value of t for 18 df at 5% level = 2.101 The calculated value of t >the table value of t. H 0 is rejected at 5% level. Therefore the marks obtained by regular students and part-time students differ significantly. Example 11 : The following data relate to the marks obtained by 11 students in two tests, one held at the beginning of the year and the other at the end of the year after intensive coaching. Do the data indicate that the students have benefited by coaching? Test 1 19 23 16 24 17 18 20 18 21 19 20 Test 2 17 24 20 24 20 22 20 20 18 22 19 Solution: Let d = x 1 – x 2 , where x 1 & x 2 are the marks in the two tests. ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 206 NOTES Anna University Chennai Test 1 = x 1 Test 2 = x 2 d = x 1 - x 2 d 2 19 17 2 4 23 24 -1 1 16 20 -4 16 24 24 0 0 17 20 -3 9 18 22 -4 16 20 20 0 0 18 20 -2 4 21 18 3 9 19 22 -3 9 20 19 1 1 -11 69 Σ d = - 11 Σ d 2 = 69 d = Σ d = -11/ 11 = -1 n s 2 = s d 2 = Σ d 2 - Σ d 2 = 69 /11 – (-1) 2 = 5.27 n n s = 2.296 If n 1 = n 2 = n and if the pairs of values of X 1 and X 2 are associated in some way or correlated we shall assume that H 0 : d (= x - y ) = 0 and test the significance of the difference between d and 0, using the test statistic t = d . with v = n- 1, s / v(n -1) where d i = x i - y i ( i = 1,2, …n ), d = x – y: and n s = S.D of d’s = 1 Σ (d i – d) 2 n i = 1 H 0 : d = 0 i.e, the students have not benefited by coaching. H 1 : d < 0 (i.e., x 1 < x 2 ) one tailed test. t = -1 . = -1.38 , v = 11 – 1 = 10 2.296 / v(10-1) | t | < 1.38 ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 207 NOTES Anna University Chennai The table value of t for 10 df at 5% level for one tailed test = The table value of t for 10 df at 10% level for two tailed test = 1.81. The calculated value of t < the table value of t. H 0 is accepted and H 1 is rejected at 5% level. Therefore there is no significant difference between the two sets of marks. i.e., the students have not benefited by coaching. How you understood ? 1. Write down the probability density of student’s t-distribution. 2. State the important properties of the t-distribution. 3. Give any two uses of t-distribution. 4. What do you mean by degrees of freedom ? 5. What is the test statistic used to the significance of the difference between the means of two small sample. TRY YOURSELF ! 1) Certain refined edible oil is packed in tins holding 16 kg each. The filling machine can maintain this but with a standard deviation of 0.5 kg. Samples of 25 are taken from the production line. If a sample mean is i) 16.35 kg ii) 15.8 kg , can we be 95% sure that the sample has come from a population of 16 kg tins? 2) A company has been producing steel tubes of mea inner diameter of 2.00 cm. A sample of 10 tubes gives an inner diameter of 2.01 cm and a variance of 0.0004 cm 2 . Is the difference in the value of mean significant? ( Value of t for 9df at 5% level = 2.262 ) 3) A random sample of 10 boys has the following IQ’s: 70, 120, 110, 101, 88, 83, 95, 98, 107, 100. Do these data support the assumption of a population mean IQ of 100 ? 4) A fertilizer mixing machine is set to give 12 kg of nitrate for every quintal bag of fertilizer. Ten 100 kg bags are examined and percentage of nitrate is s follows 11, 14, 13, 12, 13, 12, 13, 14, 11, 12. Is there reason to believe that the machine is defective? 5) Two salesman A and B are working in a certain district. From a sample survey conducted by the head office, the following results were obtained. State whether there is significant difference in the average sales between the two salesmen: DIT 111 PROBABILITY AND QUEUEING THEORY 208 NOTES Anna University Chennai A B No: of sales 20 18 Average sales (in Rs) 170 205 Standard Deviation 20 25 6) Two batches of the same product are tested for their mean life. Assuming that he life of the product follow a normal distribution with an unknown variance, test the hypothesis that the mean life is the same for both the batches, gives the following information: Batch Sample size Mean life (in hrs) SD (in hrs) I 10 750 12 II 8 820 14 7) Two sets of 10 students selected at random from a college were taken : one set was given memory test as they were and the other was given the memory test after two weeks of training and the scores are given below: Set A 10 8 7 9 8 10 9 6 7 8 Set B 12 8 8 10 8 11 9 8 9 9 Due you think there is any significant effect due to training? 8) Wire cable is manufactured by two processors. Laboratory tests were performed by putting samples of cables under tension and recording the load required ( coded units) to break the cable giving the following data. Process I 9 4 10 7 9 10 Process II 149 13 12 13 8 10 Can we say that the two processes have the same effect on the mean breaking strength, at 5% level of significance 9) A company is testing two machines. A random sample of 8 employees is selected and each employee uses each machine for one hour. The number of components produced is shown in the following table. Employee 1 2 3 4 5 6 7 8 I Machine 96 107 84 99 102 87 93 101 II Machine 99 112 90 97 108 97 94 98 DIT 111 PROBABILITY AND QUEUEING THEORY 209 NOTES Anna University Chennai Test whether there is evidence of difference between the machines in the mean number of components produced. 3.5 VARIANCE RATIO TEST OR F-TEST This test is used to test the significance of two or more sample estimates of population variance. The -statistic is defined as a ratio of unbiased estimates of population variance. Symbolically, F = S 1 2 S 2 2 where S 1 2 = Σ ( x 1 – x 1 ) 2 and S 2 2 = Σ ( x 2 – x 2 ) 2 n 1 – 1 n 2 - 1 Therefore the distribution F = S 1 2 (S 1 2 < S 2 2 ) is given by the following pdf S 2 2 f(F) = 1 (v 1 / v 2 ) v1/ 2 F (v1/ 2 - 1) . F > 0 Β (v 1 /2, v 2 /2) ( 1 + v 1 F/v 2 ) (v1 + v2 ) / 2 This is called the distribution of the variance ratio F or Senedecor’s F-distribution with v 1 = n 1 – 1 and v 2 = n 2 – 1 degrees of freedom. 3.5.1 F-test of significance of the difference between population variances and F table. If s 1 2 and s 2 2 are the variances of two samples of sizes n 1 and n 2 respectively, the estimates of the population variances based on these samples are respectively S 1 2 = n 1 s 1 2 and S 2 2 = n 2 s 2 2 n 1 -1 n 2 – 1 The quantities v 1 = n 1 – 1 and v 2 = n 2 – 1 are called the degrees of freedom of these estimates. While defining the statistic F, the larger of the two variances is always placed in the numerator and the smaller in the denominator. Senedecor has prepared tables that give, for different values of v1 and v2, the 5% and 1% critical values of F. If F denotes the observed (calculated) value and F v1,v2 (α) denotes the critical (tabulated value) of F at LOS, then P{F > F v1,v2 (α)} = α DIT 111 PROBABILITY AND QUEUEING THEORY 210 NOTES Anna University Chennai F test is not a two tailed test and is always a right tailed test , since F cannot be negative. Thus if F > F v1,v2 (α), then the difference between F and 1, i.e., the difference between S 1 2 and S 2 2 is significant at LOS ‘α’. In other words the samples may not be regarded as dawn from the sample population with the same variance. If F < F v1,v2 (α), the difference is not significant at LOS α. To test if two small samples have been drawn from the same normal population , it is not enough to test if their means differ significantly or not, because in this test we assumed that the two samples came from the same population or from populations with equal variance. So, before applying the t-test for the significance of the difference of two sample means, we should satisfy ourselves about the equality of the population variances by F-test Example 1: A sample of size 13 gave an estimated population variance of 3.0, while another sample of size 15 gave an estimate of 2.5. Could both samples be from populations with same variance? Solution: n 1 = 13 S 1 2 = 3.0 and v 1 = n 1 – 1 = 12 n 2 = 15 S 2 2 = 2.5 and v 2 = n 2 – 1 = 14 H 0 : σ 1 2 = σ 2 2 i.e the two variances have been drawn from populations with the same variance. H 1 : σ 1 2 | σ 2 2 Here S 1 2 > S 2 2 The -statistic is defined as a ratio of unbiased estimates of population variance. Symbolically, F = S 1 2 S 2 2 F = 3.0 = 1.2 2.5 v 1 = 12 and v 2 = 14 F(v 1 = 12 , v 2 = 14) at 5% LOS = 2.53 from the table value. The calculated value of F is < the tabulated value Therefore H 0 is accepted. i.e., the two samples could have come from two normal populations with the same variance. DIT 111 PROBABILITY AND QUEUEING THEORY 211 NOTES Anna University Chennai Example 2: From the following data test if the difference between the variances is significant at 5% level of significance. Sum of the squares of deviation from the mean 84.4 102.6 Size 8 10 Sample A B Solution: H 0 : σ 1 2 = σ 2 2 i.e the two variances have been drawn from populations with the same variance. H 1 : σ 1 2 | σ 2 2 Given Σ ( x 1 – x 1 ) 2 = 84.4 and Σ ( x 2 – x 2 ) 2 = 102.6 S 1 2 = Σ ( x 1 – x 1 ) 2 and S 2 2 = Σ ( x 2 – x 2 ) 2 n 1 – 1 n 2 - 1 S 1 2 = 84.4 / 7 = 12.06 and S 2 2 = 102.6/ 9 = 11.4 S 1 2 > S 2 2 The -statistic is defined as a ratio of unbiased estimates of population variance. Symbolically, F = S 1 2 S 2 2 F = 12.06 = 1.058 11.4 v 1 = n 1 – 1 = 7 and v 2 = n 2 -1 = 9 F(v 1 = 7 , v 2 = 9) at 5% LOS = 3.29 from the table value. The calculated value of F is < the tabulated value Therefore H 0 is accepted. i.e., the two samples could have come from two normal populations with the same variance. Example 3: Time taken by workers in performing a job are given below: – Method 1 20 16 26 27 23 22 Method 2 27 33 42 35 32 34 38 Test whether there is any significant difference between the variances of time distribution. DIT 111 PROBABILITY AND QUEUEING THEORY 212 NOTES Anna University Chennai Solution: Let us first calculate the variance of the samples. Sample I Sample II x-22 y-34 x d d 2 y d d 2 20 -2 4 27 -7 49 16 -6 36 33 -1 1 26 4 16 42 8 64 27 5 25 35 1 1 23 1 1 32 -2 4 22 0 0 34 0 0 38 4 16 134 2 82 241 3 135 n 1 = 6 n 2 = 7 x = Σ x / n 1 = 134 / 6 = 22.33 y = Σ y / n 2 = 241 / 7 = 34.43 Let d 1 = x – 22 d 2 = y – 34 s 1 2 = Σ d 1 2 - Σ d 1 2 n 1 n 1 s 1 2 = 82 / 6 – ( 2 / 6 ) 2 = 13.67 – 0.44 = 13.23 s 2 2 = Σ d 2 2 - Σ d 2 2 n 2 n 2 s 2 2 = 135 / 7 –(3 / 7) 2 = 19.29 – 0.18 = 19.11 S 1 2 = n 1 s 1 2 = 6 × 13.23 = 15.88 and S 2 2 = n 2 s 2 2 = 7 × 19.11 = 22.3 n 1 -1 5 n 2 – 1 6 Here S 2 2 > S 1 2 H 0 : σ 1 2 = σ 2 2 i.e the two variances have been drawn from populations with the same variance. H 1 : σ 1 2 | σ 2 2 ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 213 NOTES Anna University Chennai The -statistic is defined as a ratio of unbiased estimates of population variance. Symbolically, F = S 2 2 S 1 2 F = 22.30 = 1.40 15.88 v 1 = n 1 – 1 = 6and v 2 = n 2 -1 = 5 F(v 1 = 6 , v 2 = 5) at 5% LOS = 4.28 from the table value. The calculated value of F is < the tabulated value Therefore H 0 is accepted. i.e., the two samples could have come from two normal populations with the same variance. Example 4: Two random samples drawn from normal population are – Sample I 20 16 26 27 23 22 18 24 25 19 Sample II 27 33 42 35 32 34 38 28 41 43 30 37 Obtain estimates of variances of the population and test whether the two populations have the same variances. Solution: Let us first calculate the variance of the samples. Sample I Sample II x-22 y-35 x d d 2 y d d 2 20 -2 4 27 -8 64 16 -6 36 33 -2 4 26 4 16 42 7 49 27 5 25 35 0 0 23 1 1 32 -3 9 22 0 0 34 -1 1 18 -4 16 38 3 9 24 2 4 28 -7 49 25 3 9 41 6 36 19 -3 9 43 8 64 30 -5 25 37 2 4 220 0 120 420 0 314 DIT 111 PROBABILITY AND QUEUEING THEORY 214 NOTES Anna University Chennai n 1 = 10 n 2 = 12 x = Σ x / n 1 = 220 / 10 = 22 y = Σ y / n 2 = 420 / 12 = 35 Let d 1 = x – 22 d 2 = y – 35 s 1 2 = Σ d 1 2 - Σ d 1 2 n 1 n 1 s 1 2 = 120 / 10 – 0 = 12 s 2 2 = Σ d 2 2 - Σ d 2 2 n 2 n 2 s 2 2 = 314 / 12 - 0 = 26.17 S 1 2 = n 1 s 1 2 = 10 × 12 = 13.33 and S 2 2 = n 2 s 2 2 = 12 × 26.17 = 28.55 n 1 -1 9 n 2 – 1 11 Here S 2 2 > S 1 2 H 0 : σ 1 2 = σ 2 2 i.e the two variances have been drawn from populations with the same variance. H 1 : σ 1 2 | σ 2 2 The -statistic is defined as a ratio of unbiased estimates of population variance. Symbolically, F = S 2 2 S 1 2 F = 28.55 = 2.14 13.33 v 1 = n 1 – 1 = 11and v 2 = n 2 -1 = 9 F(v 1 = 11, v 2 = 9) at 5% LOS = 3.10 from the table value. The calculated value of F is < the tabulated value Therefore H 0 is accepted. i.e., the two samples could have come from two normal populations with the same variance. ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 215 NOTES Anna University Chennai Example 5: Values of a variate in two samples are given below – Sample I 5 6 8 1 12 4 3 9 6 10 Sample II 2 3 6 8 1 10 2 8 Test the significance of the difference between the two sample means and the two sample variances. Solution: Let us first calculate the variance of the samples. Sample I Sample II x x 2 y y 2 5 25 2 4 6 36 3 9 8 64 6 36 1 1 8 64 12 144 1 1 4 16 10 100 3 9 2 4 9 81 8 64 6 36 64 10 100 64 512 40 282 n 1 = 10 n 2 = 8 x = Σ x / n 1 = 64 / 10 = 6.4 y = Σ y / n 2 = 40 / 8 = 5 Let d 1 = x – 22 d 2 = y – 35 s 1 2 = Σ x 1 2 - Σ x 1 2 n 1 n 1 s 1 2 = 512 / 10 – ( 64 / 10) 2 = 51.2 – 40.96 = 10.24 ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 216 NOTES Anna University Chennai s 2 2 = Σ x 2 2 - Σx 2 2 n 2 n 2 s 2 2 = 282 / 8 – ( 40 / 8) 2 = 35.25 – 25 = 10.25 case 1: Test for mean: The samples are small and so we apply t-test. H 0 :µ 1 = µ 2 (The means of the two population are equal) H 1 : µ 1 | µ 2 (The means of the two population are not equal) The test statistic is given by t = x 1 – x 2 . S 1 + 1 n 1 n 2 where S 2 = n 1 s 1 2 + n 2 s 2 2 n 1 + n 2 – 2 n 1 + n 2 – 2 is the number of degrees of freedom of the statistic. S 2 = 10 × 10.24 + 8 × 10.25 = 11.525 10 + 8 – 2 S = 3.395 Therefore t = 6.45 - 5 . = 0.87 3.395 \( 1/ 10 + 1/8) |t| = 0.87 number of degrees of freedom, ndf = 10 + 8– 2 = 16 The table value of t for 16 df at 5% level = 2.12 The calculated value of t < the table value of t. H 0 is accepted at 5% level. Hence there is no significant difference between the means of the population. Case 2:Test for variance n 1 = 10 n 2 = 8 s 1 2 = 10.24 s 2 2 = 10.25 S 1 2 = n 1 s 1 2 = 10 × 10.24 = 11.38 and S 2 2 = n 2 s 2 2 = 8 × 10.25 = 11.71 n 1 -1 9 n 2 – 1 7 ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 217 NOTES Anna University Chennai Here S 2 2 > S 1 2 H 0 : σ 1 2 = σ 2 2 i.e the two variances have been drawn from populations with the same variance. H 1 : σ 1 2 | σ 2 2 The -statistic is defined as a ratio of unbiased estimates of population variance. Symbolically, F = S 2 2 S 1 2 F = 11.71 = 1.03 11.38 v 1 = n 1 – 1 = 7and v 2 = n 2 -1 = 9 F(v 1 = 11, v 2 = 9) at 5% LOS = 3.29 from the table value. The calculated value of F is < the tabulated value Therefore H 0 is accepted. i.e., the two samples could have come from two normal populations with the same variance. Example 6 : Two random samples gave the following data – Size Mean Variance Sample I 8 9.6 1.2 Sample II 11 16.5 2.5 Can we conclude that two samples have been drawn from the same normal population. Solution: To conclude that the two samples have been drawn from the same population, we have to check first that the variances of the populations do not differ significantly and then check that the sample means (and hence the population means) do not differ significantly n 1 = 10 n 2 = 8 x 1 = 9.6 x 2 = 16.5 s 1 2 = 1.2 s 2 2 = 2.5 DIT 111 PROBABILITY AND QUEUEING THEORY 218 NOTES Anna University Chennai Case 1:Test for variance S 1 2 = n 1 s 1 2 = 8 × 1.2 = 1.37 and S 2 2 = n 2 s 2 2 = 11 × 2.5 = 2.75 n 1 -1 7 n 2 – 1 10 H 0 : σ 1 2 = σ 2 2 i.e the two variances have been drawn from populations with the same variance. H 1 : σ 1 2 | σ 2 2 The -statistic is defined as a ratio of unbiased estimates of population variance. Symbolically, F = S 2 2 S 1 2 F = 2.75 = 2.007 1.37 v 1 = n 1 – 1 = 10and v 2 = n 2 -1 = 7 F(v 1 = 10, v 2 = 7) at 5% LOS = 3.64 from the table value. The calculated value of F is < the tabulated value Therefore H 0 is accepted. i.e., the two samples could have come from two normal populations with the same variance. Case 2: Test for mean: The samples are small and so we apply t-test. H 0 :µ 1 = µ 2 (The means of the two population are equal) H 1 : µ 1 | µ 2 (The means of the two population are not equal) The test statistic is given by t = x 1 – x 2 . S 1 + 1 n 1 n 2 where S 2 = n 1 s 1 2 + n 2 s 2 2 n 1 + n 2 – 2 n 1 + n 2 – 2 is the number of degrees of freedom of the statistic. S 2 = 8 × 1.2 + 11 × 2.5 = 11.24 17 S = 2.1823 Therefore t = 9.6 - 16.5 . = -10.05 2.1823 \( 1/ 8+ 1/ 11 ) DIT 111 PROBABILITY AND QUEUEING THEORY 219 NOTES Anna University Chennai |t| = 10.05 number of degrees of freedom, ndf = 8 + 11 – 2 = 17 The table value of t for 17 df at 5% level = 2.11 The calculated value of t > the table value of t. H 0 is rejected at 5% level. Hence there is significant difference between the means of the population. Hence the two samples could not have been drawn from the same normal population. How you understood ? 1.State the important properties of the F-distribution. 2.What is the use of F-distribution? 3.Write down the probability density function of the F-distribution. TRY YOURSELF ! 1) In a sample of 8 observations, the sum of the squared deviations of items from the Mean was 94.5. In another sample of 10 observations, the value was found to be 101.7. Test whether the difference in the variances is significant at 5% level. 2) Two samples were drawn from two normal populations and their values are A 66 67 75 76 82 84 88 90 92 B 64 66 74 78 82 85 87 92 93 95 97 Test whether the two populations have the same variance at 5% level of significance. 3) In tests given to two groups of students drawn from two different populations, the marks obtained were as follows Group A 18 20 36 50 49 36 34 49 41 Group B 29 28 26 35 30 44 46 DIT 111 PROBABILITY AND QUEUEING THEORY 220 NOTES Anna University Chennai 3.6 CHI SQUARE TEST Karl Pearson has shown that if X 1 , X 2 , X 3 , . . . X n are n independent normal variables with means µ 1 ,µ 2 ,. . . .µn and standard deviation σ 1 ,σ 2 , . . . . σ n then the random variable defined by χ 2 = X 1 2 + X 2 2 +. . . .+X n 2 has a probability distribution called χ 2 distribution with n degrees of freedom. Here n is only the number of independent variables under consideration. The importance of this distribution is that obeys additive property.  ADDITIVE PROPERTY If χ 1 2 ,χ 2 2 , . . . . χ k 2 are k independent χ 2 random variables with n 1 ,n 2 ,…,n k degrees of freedom then their sum χ 2 = χ 1 2 + χ 2 2 + . . . .+ χ k 2 is also a χ 2 random variable with n 1 + n 2 + … + n k number of degrees of freedom. Pearson has shown that χ 2 – statistic is useful for comparison of observed frequencies with theoretical frequencies and to draw the decision whether there is any significant difference between these two sets. In this context χ 2 is called a non- parametric test.  Pearson’s Statistics For testing the significance of difference between observed and expected frequencies under the null hypothesis that the difference is insignificant, Pearson has constructed the statistic that χ 2 = Σ (O i – E i ) 2 E i Here O i are the observed frequencies and E i are the expected frequencies. The expected frequencies can be calculated on the assumption of H 0 . Pearson has shown that for large sample, this statistic follows χ 2 distribution with n-1 degrees of freedom. The sampling distribution of χ 2 is given by f(χ 2 ) = c(χ 2 ) n/2 – 1 exp(-χ n/2 ) where the constant c is to be determined such that I f(χ 2 )d χ 2 = 1 The χ 2 distribution has only one parameter v called the number of degrees of freedom. For each value of v χ 2 has different curve. For small values of v the curve is skewed the the right. For large values of v, χ 2 distribution is closed approximated to the normal distribution. [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 221 NOTES Anna University Chennai  USES OF χ 2 TEST The following are the uses of χ 2 statistic – (1) It is used to test the goodness of fit of a distribution. (2) It is used to test the significance of the difference between the observed frequencies in a sample and the expected frequencies obtained from the theoretical distribution. (3) It is used to test the independence of the attributes. (4) In the case of small samples ( where the population standard deviation is not known), χ 2 statistic is used to test whether a specified value can be the population variance σ 2 . 3.6.1 χ 2 Test for Goodness of Fit Procedure for testing the significance of the difference between the observed and expected frequencies. H 0 – There is no significant differences between the observed and the expected frequencies. H 1 – There is significant difference between the observed and the expected frequencies. The test statistic is χ 2 = Σ (O – E) 2 E the expected frequencies are determined on the assumption that H 0 is true. The number of degrees of freedom = n-1 where n is the number of classes. From the χ 2 table we can find for a given degrees of freedom the table value of χ 2 for a given significance level ( say α = .05 or α = 0.01 ) If the calculated value of χ 2 < the table value of χ 2 , H 0 is accepted at the significance level α. If the calculated value of χ 2 > the table value of χ 2 , H 0 is rejected at the significance level α. Note - For testing the goodness of fit of a distribution by assuming H 0 as some specific distribution, Binomial, Poisson etc, we calculate the theoretical frequencies and adopt the procedure given above to test whether the assumed distribution is a better fit for the observed frequencies. DIT 111 PROBABILITY AND QUEUEING THEORY 222 NOTES Anna University Chennai  Conditions for the validity of χ 2 test 1. The number of observations N in the sample must be reasonably large, say ¡ 50 2. Individual frequencies must not be too small, i.e., O ¡ 10. In case of O , 10, it is combined with the neighbuoring frequencies, so that the combined frequency is ¡ 10. 3. The number of classes n must be neither too small nor too large i.e., 4 , n , 16. 3.6.2 Test of independence of attributes Another important application of the χ 2 distribution is the testing of independence of attributes ( attributes are characters which are non measurable – for eg. Sex, Employment, Literacy etc are all attributes ). Suppose we want to test whether sex and employment are associated. In this case take a random sample from the population an classify the sample as given in the following table. The numbers in the table denote the frequencies ( number of persons possessing the attribute ) Male Female Total Employed 50 20 70 Unemployed 15 15 30 Total 65 35 100 This type of table which has one basis of classification across column and another across row is known as contingency table. The above table has 2 rows and 2 columns and hence is called as 2 X 2 contingency table. A table which as r rows and s columns is called a r X s contingency table. In testing the hypothesis the null hypothesis is taken as “ employment is independent of sex” where as the alternate hypothesis is “ employment is not independent of sex”. Then comes the question of determining the expected frequencies. Assuming that H 0 is true, the totals are all kept the same. For example, the expected frequency for the 1 st cell in the above table, is determined by the formula : Row total × Column total Grand Total = 70 X 65 = 45.5 100 DIT 111 PROBABILITY AND QUEUEING THEORY 223 NOTES Anna University Chennai The other theoretical frequencies are determined on the same lines – Male Female Total Employed 45.5 24.5 70 Unemployed 24.5 5.5 30 Total 65 35 100 It can be checked that by determining the only one cell frequency the other expected frequencies can be easily obtained from the column and row totals. Thus in a 2 X 2 contingency table the number of degrees of freedom is ( 2 – 1 )X(2 – 1 ) = 1. In general in a r X s contingency table the number of degrees of freedom is ( r – 1 ) X ( s – 1 ). Test procedure – Step 1 – write down the null hypothesis. Step 2 – write down the alternate hypothesis. Step 3 – calculate the theoritial frequencies for the contingency table. Step 4 – calculate χ 2 = Σ (O – E) 2 E Step 5 – write down the number of degrees of freedom. Step 6 - draw the conclusion on the hypothesis by comparing the calculated values of χ 2 with the table value of χ 2 . Note the value of χ 2 statistic for a 2 X 2 contingency table can also be calculated using the formula given below – A Α Total B a b a + b Β c d c + d Total a + c b + d N χ 2 = N(ad - bc) 2 ( a + b)( c + d )( a + c )( a + d) 3.6.3 Test for a specified population variance Let {x 1 ,x 2 , … x n } be a random sample of size n drawn from a normal population. We want to test, based on the sample, whether the population variance can be σ 0 2 . Let us now give the procedure for the test H 0 : σ 2 = σ 0 2 . DIT 111 PROBABILITY AND QUEUEING THEORY 224 NOTES Anna University Chennai H 1 : σ 2 | σ 0 2 . The test statistic is χ 2 . On the assumption that H 0 is true, it has been shown that the statistic χ 2 = ns 2 / σ 2 has a χ 2 distribution with (n - 1) degrees of freedom. In this formula n is the sample size , s 2 is the variance, σ 2 is the population variance We can determine the table value of χ 2 for (n - 1) degrees of freedom. Accept H 0 if the calculated value of χ 2 < the table value. Reject H 0 if the calculated value of χ 2 > the table value. Example 1: A company keeps records of accidents. During a recent safety review, a random sample of 60 accidents was selected and classified by the day of the week on which they occurred. Day Mon Tue Wed Thu Fri No. of Accidents 8 12 9 14 17 Test whether there is any evidence that the accidents are more likely on some days than others. Solution: H 0 – Accidents are equally likely to occur on nay day of the week. H 1 – Accidents are not equally likely to occur on the days of the week. Total number of accidents = 60 On the assumption H 0, the expected number of accidents on any day = 60 = 12 5 Let O denote the observed frequency and E denote the expected frequency O E‘ O - E ( O - E ) 2 8 12 -4 16 12 12 0 0 9 12 -3 9 14 12 2 4 17 12 5 25 60 60 54 DIT 111 PROBABILITY AND QUEUEING THEORY 225 NOTES Anna University Chennai χ 2 = Σ (O – E) 2 E = 54 = 4.5 12 N = number of classes = 5 Thus number of degrees of freedom = n – 1 = 5 – 1 = 4 For 4 degrees of freedom the table value of χ 2 is 9.4888. But the calculated value of χ 2 is 4.5 Thus the calculated value of χ 2 < the table value of χ 2 Hence H 0 is accepted at 5% level. This means that accidents are equally likely to occur on any day of the week. Example 2: A company produces a product of 4 sizes : small, medium, large and extra large. In the past the demand for these sizes has been fairly constant at 20% for small, 45% for medium, 25% for large and 10% for extra large. A random sample of 400 recent sales included 66 small, 172 medium, 109 large and 53 extra large. Test whether there is evidence of significant change in demand for the different sizes. Solution: H 0 – There is no evidence of significant change in demand for the different sizes. H 1 – There is evidence of significant change in demand for the different sizes. The expected frequencies are – 20 x 400, 45 x 400, 25 x 400 and 10 x 400 100 100 100 100 i.e., 80,180,100,40 Size O E‘ O - E ( O - E ) 2 ( O – E ) 2 E Small 66 80 -14 196 2.45 Medium 172 180 -8 64 0.356 Large 109 100 9 81 0.810 Extra Large 53 40 13 169 4.225 400 7.841 [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 226 NOTES Anna University Chennai χ 2 = Σ (O – E) 2 = 7.841 E Number of degrees of freedom = 4 – 1 = 3 For 3 degrees of freedom the table value of χ 2 at 5% level is 7.81. But the calculated value of χ 2 is 7.841 Thus the calculated value of χ 2 > the table value of χ 2 Hence H 0 is rejected at 5% level. This means that there is evidence of significant change in demand for the different sizes. Example 3: In 20 throws of a single die the following distributions of faces was observed Face 1 2 3 4 5 6 Frequency 30 25 18 10 22 15 Can you say that the die is unbiased? Solution: H 0 – The die is unbiased. H 1 – The die is biased. On the assumption H 0 , the expected frequency for each face = 120 x 1 = 20 6 Face O E O - E ( O - E ) 2 1 30 20 10 100 2 25 20 5 25 3 18 20 -2 4 4 10 20 -10 100 5 22 20 2 4 6 15 20 -5 25 120 258 If E is same for all u need not have a separate column to find ( O – E ) 2 E χ 2 = Σ (O – E) 2 = 258 = 12.9 E 20 [ ] [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 227 NOTES Anna University Chennai Number of degrees of freedom = n – 1 = 6 – 1 = 5 For 5 degrees of freedom the table value of χ 2 at 5% level is 11.07 But the calculated value of χ 2 is 12.9 Thus the calculated value of χ 2 > the table value of χ 2 Hence H 0 is rejected at 5% level. Hence the die can be regarded as biased. Example 4: A sample analysis of examination results of 500 students was made. It was found that 220 students have failed, 170have secured a third class, 90 have secured a second class, and the rest, a first class. Do these figures support the general brief that the above categories are in the ratio 4:3:2:1 respectively? Solution: H 0 – The results in the four categories are in the ratio 4:3:2:1. H 1 – The results in the four categories are not in the ratio 4:3:2:1. On the assumption H 0, t he expected frequencies are – 40 x 500, 3 x 500, 2 x 500 and 1 x 500 10 10 10 10 i.e., 200, 150, 100, 50. O E O - E ( O - E ) 2 ( O – E ) 2 E Failures 220 200 20 400 2.000 III class 170 150 20 400 2.667 II class 90 100 -10 100 1.000 I class 20 50 -30 900 18.000 500 23.667 χ 2 = Σ (O – E) 2 = 23.667 E Number of degrees of freedom = 4 – 1 = 3 For 3 degrees of freedom the table value of χ 2 at 5% level is 7.81. But the calculated value of χ 2 is 23.667 Thus the calculated value of χ 2 > the table value of χ 2 [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 228 NOTES Anna University Chennai Hence H 0 is rejected at 5% level. This means the results in the four categories are not in the ratio 4:3:2:1. Example 5: The following table shows the distribution of digits in numbers chosen at random from a telephone directory Digit 0 1 2 3 4 5 6 7 8 9 Frequency 1026 1107 997 966 1075 933 1107 972 964 853 Test whether the digits may be taken t occur equally frequently in the directory. Solution: H 0 – The digits occur equally frequently in the directory. H 1 – The digits do not occur equally frequently On the assumption H 0 , the expected frequency for each face = 10000 = 1000 10 Digit O E O - E ( O - E ) 2 0 1026 1000 26 0.676 1 1107 1000 107 11.449 2 997 1000 3 0.009 3 966 1000 34 1.156 4 1075 1000 75 5.625 5 933 1000 67 4.489 6 1107 1000 107 11.449 7 972 1000 28 0.784 8 964 1000 36 1.296 9 853 1000 147 21.609 10000 120 58.542 If E is same for all u need not have a separate column to find ( O – E ) 2 E χ 2 = Σ (O – E) 2 = 58.542 E Number of degrees of freedom = n – 1 = 10 – 1 = 5 For 5 degrees of freedom the table value of χ 2 at 5% level is 16.919 But the calculated value of χ 2 is 58.542. [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 229 NOTES Anna University Chennai Thus the calculated value of χ 2 > the table value of χ 2 Hence H 0 is rejected at 5% level. The digits are not uniformly distributed in the directory. Example 6: A set of 5 identical coins is tossed 320 times and the number of heads appearing each time is recorded. 0 1 2 3 4 5 14 45 80 772 61 8 Test whether the coins are unbiased at 5% level of significance. Solution: H 0 : coins are unbiased (P(getting head ) = p = ½ , q = 1/6 ) H 1 coins are not biased On the assumption H 0 , the probability of getting exactly ‘r’ successes = 5Crp r q 5 - r (r = 0, 1, 2, …5) Therefore the expected number of times in which exactly ‘r’ successes are obtained = 320 ×5C r p r q 5 - r = 10, 50, 100, 100, 5, 10 No: of O E O - E ( O – E ) 2 ( O – E ) 2 heads E 0 14 10 4 16 1.60 1 45 50 -5 25 0.50 2 80 100 -20 400 4.0 3 112 100 12 144 1.44 4 1 5 11 121 2.42 5 8 10 -2 4 0.40 320 320 10.36 χ 2 = ( O – E ) 2 = 10.36 E Number of degrees of freedom = n – 1 = 6 – 1 = 5 Table value of χ 2 for 5 at 5% level = 11.07 Since the calculated value of χ 2 is less than the table value of χ 2 , H 0 is accepted at 5% level. [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 230 NOTES Anna University Chennai Hence the coins are unbiased. Example 7: A survey of 320 families with five children each revealed the following distribution. No: of boys 0 1 2 3 4 5 No: of girls 5 4 3 2 1 0 No: of families 12 40 88 110 56 14 Is the result consistent with the hypothesis that male and female births are equally probable? Solution: H 0 : male and female births are equally probable (P(male birth ) = p = ½ , q = 1/6 ) H 1 : male and female births are not equally probable On the assumption H 0 , the probability that a family of 5 children has r male children = 5Crp r q 5 – r (r = 0, 1, 2, …5) Therefore the expected number of times in which exactly ‘r’ successes are obtained = 320 ×5C r p r q 5 - r = 10, 50, 100, 100, 5, 10 No: of males O E O - E ( O – E ) 2 ( O – E ) 2 E 0 12 10 2 4 0.4 1 40 50 -10 100 0.5 2 88 100 -22 484 4.84 3 110 100 10 100 1 4 56 50 6 36 0.72 5 14 10 4 16 1.6 320 320 7.16 χ 2 = ( O – E ) 2 = 7.16 E Number of degrees of freedom = n – 1 = 6 – 1 = 5 Table value of χ 2 for 5 at 5% level = 11.07 Since the calculated value of χ 2 is less than the table value of χ 2 , H 0 is accepted at 5% level. Hence the male and female births are equally probable. [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 231 NOTES Anna University Chennai Example 8: Fit a binomial distribution for the following data and also test the goodness of fit x 0 1 2 3 4 5 6 Total f 5 18 28 12 7 6 4 80 Solution: H 0 : the given distribution is approximately a binomial distribution To find the binomial distribution N(q + p) n , which fits the given data, we require p. We know that the mean of the binomial distribution is np, from which we can find p. Now the mean of the given distribution is found out and is equated to np. x 0 1 2 3 4 5 6 Total f 5 18 28 12 7 6 4 80 fx 0 18 56 36 28 30 24 192 x = Σ fx = 192/80 = 2.4 Σ f i.e., np = 2.4 or 6p = 2.4, since the maximum value taken by x is n. p = 0.4 and hence q = 0.6 The expected frequencies are given by = 80 ×6C r p r q 6 – r ( r = 0, 1, 2, 3, 6) = 3.73 [ 4, 14.93[15, 24.88 [25, 22.12[ 22, 11.06 [ 11, 2.95 [ 3, 0.33 [ 0 O 5 18 28 12 7 6 4 E 4 15 25 22 11 3 0 The first class is combined with the second and the last two classes are combined with the last but second class in order to make the expected frequency in each class greater than or equal to 10. Thus after grouping O 23 28 12 17 E 19 25 22 14 DIT 111 PROBABILITY AND QUEUEING THEORY 232 NOTES Anna University Chennai O E O - E ( O – E ) 2 ( O – E ) 2 E 23 19 4 16 0.8421 28 25 3 9 0.36 12 22 10 100 0.5455 17 14 3 9 0.6429 6.39 χ 2 = Σ ( O – E ) 2 = 6.39 E Number of degrees of freedom = n – k = 4 – 2 = 2 Table value of χ 2 for 2at 5% level = 5.99 Since the calculated value of χ 2 is > the table value of χ 2 , H 0 is rejected at 5% level. i.e., the binomial fit for the given distribution is not satisfactory. Example 9: Fit a Poisson distribution for the following data and also test the goodness of fit x 0 1 2 3 4 5 Total f 142 156 69 27 5 1 400 Solution: H 0 : the given distribution is approximately a Poisson distribution To find the Poisson distribution whose probability law is P(X = r) = e -λ λ r , r = 0, 1, 2, . … r! We require λ the mean of the Poisson distribution. We will find the mean of the given data and assume it as λ. x 0 1 2 3 4 5 Total f 142 156 69 27 5 1 400 fx 0 156 138 81 20 5 400 x = Σ fx = 400/400 = 1 Σ f [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 233 NOTES Anna University Chennai The expected frequencies are given by Ne -λ λ r , r = 0, 1, 2, . …= > 400e -λ λ r , r = 0, 1, 2, . r! r! = 147.15 [ 147, 147.15 [ 147, 73.58 [ 74 , 24.53 [ 25, 6.13 [ 6, 1.23 [ 1 O 142 156 69 27 5 1 E 147 147 74 25 6 1 The last three classes are combined into one, so that the expected frequency in the class may be greater than 10. Thus after regrouping, we have O 142 156 69 33 E 147 147 74 32 O E O - E ( O – E ) 2 ( O – E ) 2 E 142 147 -5 25 0.17 156 147 9 81 0.551 69 74 -5 25 0.027 33 32 1 1 0.0312 1.09 χ 2 = ( O – E ) 2 = 1.09 E Number of degrees of freedom = n – k = 4 – 2 = 2 Table value of χ 2 for 2at 5% level = 5.99 Since the calculated value of χ 2 is < the table value of χ 2 , H 0 is accepted at 5% level. i.e., the Poisson fit for the given distribution is satisfactory. Problems on Independence of attributes Example 10 : A random sample of employees of a large company was selected and the employees were asked to complete a questionnaire. One question asked was whether the employee was in favour of the introduction of flexible working hours. The following table classifies the employees by their response and by their area of work. [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 234 NOTES Anna University Chennai Area of work Response Production Non Production In favour 129 171 Not in favour 1 69 Test whether there is evidence of a significant association between the response and the area of work? Solution: H 0 : There is no evidence of a significant association between the response and the area of work H 1 : There is an evidence of a significant association between the response and the area of work Now we have to calculate the expected frequencies to apply the χ 2 test. On the assumption of H 0 , the expected frequency for the class ‘’production an in favour” is given by (A ) x (B) = 160 x 300 = 120 N 400 Similarly we can calculate the other expected frequencies . The other expected frequencies are 240 x 300 = 180, 160 x 100 = 40, 240 x 100 = 60 400 400 400 Table showing observed frequencies Response Production Non Production Total In favour 129 171 300 Not in favour 1 69 100 Total 160 240 400 Table showing Expected frequencies DIT 111 PROBABILITY AND QUEUEING THEORY 235 NOTES Anna University Chennai Response Production Non Production Total In favour 129 180 300 Not in favour 40 60 100 Total 160 240 400 O E O - E ( O – E ) 2 ( O – E ) 2 E 129 120 9 81 0.675 171 180 -9 81 .450 31 40 -9 81 2.025 69 60 9 81 1.350 400 400 4.500 χ 2 = ( O – E ) 2 = 4.5 E Number of degrees of freedom = (2 – 1)(2 - 1) = 1 Table value of χ 2 for 1 at 5% level = 3.81 Since the calculated value of χ 2 is greater than the table value of χ 2 , H 0 is rejected at 5% level. Hence there is evidence for a significant association between response and the area of work Example 11 : Can vaccination be regarded as a preventive measure of small-pox evidenced by the following data? ‘’Of 1482 persons exposed to small-pox in a locality 368 in all were attacked”. Given the chi-square value at 5% level of significance fro 1 df is 3.84 Solution: H 0 : There is no evidence that vaccination can be regarded as a preventive measure of small-pox H 1 : There is evidence that vaccination can be regarded as a preventive measure of small-pox [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 236 NOTES Anna University Chennai Table showing observed frequencies Vaccinated Non Vaccinated Total Attacked 35 333 368 Not Attacked 308 806 1114 Total 343 1139 1482 Table showing Expected frequencies Response Vaccinated Non Vaccinated Total Attacked 343 x 368 = 85 283 368 1482 Not Attacked 258 856 1114 Total 343 1139 400 χ 2 = ( O – E ) 2 = 50.85 E Number of degrees of freedom = (2 – 1)(2 - 1) = 1 Table value of χ 2 for 1 at 5% level = 3.81 Since the calculated value of χ 2 is greater than the table value of χ 2 , H 0 is rejected at 5% level. Hence there is evidence for regarding vaccination as a preventive measure for small- pox. Example 12: To test the efficiency of a new drug a controlled experiment was conducted wherein 300 patients were administered the new drug and 200 other patients were not given the drug. The patients were monitored and the results were obtained as follows: Cured Condition No effect worsened Given the drug 200 40 60 Not given the drug 120 30 50 Use χ 2 test for finding the effect of the drug. [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 237 NOTES Anna University Chennai Solution: H 0 : The drug is not effective H 1 : The drug is effective Table showing observed frequencies Cured Condition No effect Total worsened Given the drug 200 40 60 300 Not given the drug 120 30 50 200 320 70 110 500 Table showing Expected frequencies Cured Condition No effect Total worsened Given the drug 320 x 300 = 192 42 66 300 500 Not given the drug 128 28 44 200 320 70 110 500 O E O - E ( O – E ) 2 ( O – E ) 2 E 200 192 8 64 0.3313 40 42 -2 4 0.0952 60 66 -6 36 0.5454 120 128 -8 64 0.5000 30 28 2 4 0.1429 50 44 6 36 0.8182 500 500 2.4330 χ 2 = ( O – E ) 2 = 2.43 E [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 238 NOTES Anna University Chennai Number of degrees of freedom = (2 – 1)(3 - 1) = 2 Table value of χ 2 for 2 at 5% level = 5.991 Since the calculated value of χ 2 is < the table value of χ 2 , H 0 is accepted at 5% level. Hence the drug is not effective. Example 13 : A sample of hotels in a particular country was selected. The following table shows the number of hotels in each region of t h country and in each of four grades Grade Region Eastern central Western 1 star 29 22 29 2 star 67 38 55 3 star 53 32 35 4 star 11 8. 21 Show that there is evidence of a significant association between region and grade of hotel in this country. Solution: H 0 : There is no evidence for significant association between region and grade of hotel H 1 : There is evidence for significant association between region and grade of hotel Table showing observed frequencies Region Total Eastern central Western 1 star 29 22 29 80 2 star 67 38 55 160 3 star 53 32 35 120 4 star 11 8. 21 40 160 100 140 400 Table showing Expected frequencies Region Total Eastern central Western 1 star 32 0 28 80 2 star 64 60 56 160 3 star 48 30 42 120 4 star 16 10 14 40 160 100 140 400 DIT 111 PROBABILITY AND QUEUEING THEORY 239 NOTES Anna University Chennai O E O - E ( O – E ) 2 ( O – E ) 2 E 29 32 -3 9 0.281 22 20 2 4 0.200 29 28 1 1 0.036 67 64 3 9 0.141 38 40 -2 4 0.200 5 56 -1 1 0.018 53 48 5 25 0.521 32 30 2 4 0.133 35 42 -7 49 1.167 11 16 -5 25 1.562 8 10 -2 4 0.40 1 14 7 49 3.500 400 400 8.519 χ 2 = ( O – E ) 2 = 8.159 E Number of degrees of freedom = (4 – 1)(3 - 1) = 6 Table value of χ 2 for 6 at 5% level = 12.59 Since the calculated value of χ 2 is > the table value of χ 2 , H 0 is rejected at 5% level. Hence there is evidence for significant association between region and grade of hotel. Example 14 :A credit rating agency conducted a survey of customers and analyses them by occupation and credit risk. The results were as follows: Credit rating Administrative Skilled manual Semi-skilled & clerical & unskilled High 60 50 10 Average 30 20 10 Poor 10 10 40 Test whether there is any association between occupation and credit rating? Solution: H 0 : There is no association between occupation and credit rating H 1 : There is association between occupation and credit rating [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 240 NOTES Anna University Chennai Table showing observed frequencies Credit rating Administrative Skilled manual Semi-skilled Total & clerical & unskilled High 60 50 10 120 Average 30 20 10 60 Poor 10 10 40 60 100 80 60 240 Table showing Expected frequencies Credit rating Administrative Skilled manual Semi-skilled Total & clerical & unskilled High 100 x 120 = 50 40 30 120 240 Average 25 20 15 60 Poor 25 20 15 60 100 80 60 240 O E O - E ( O – E ) 2 ( O – E ) 2 E 60 50 10 100 2.00 50 40 10 100 2.50 10 30 -20 400 13.33 30 25 5 25 1.00 20 20 0 0 0.00 10 15 -5 25 1.67 10 25 -15 25 9.00 10 20 -10 100 5.00 40 15 25 625 41.67 400 400 76.17 χ 2 = ( O – E ) 2 = 76.17 E Number of degrees of freedom = (3 – 1)(3 - 1) = 4 Table value of χ 2 for 4 at 5% level = 9.49 [ ] DIT 111 PROBABILITY AND QUEUEING THEORY 241 NOTES Anna University Chennai Since the calculated value of χ 2 is > the table value of χ 2 , H 0 is rejected at 5% level. Therefore there is association between occupation and credit rating . Problems on test of specified population variance Example 15: Weights in Kg of 10 students are given below: 38,40,45, 53, 47, 43, 55, 48, 52, 49 Can we say that the variance of the distribution of weights of all students from which the above sample of 10 students was drawn is equal to20 square kg? Solution: Here we have to apply the χ 2 –test for testing the significance of the difference between the sample variance and the population variance. H 0 : σ 2 = 20 (there is no significant difference between the sample variance and the population variance) H 1 : σ 2 | 20 (there is significant difference between the sample variance and the population variance) x d d 2 38 -9 81 40 -7 49 45 -2 4 53 6 36 47 0 0 43 -4 16 55 8 4 48 1 1 52 5 25 49 2 4 470 280 d = x – 47 s 2 = d 2 - d 2 = 280 - 0 = 28 kg 2 n n 10 σ 2 = 20 kg 2 n = 10 χ 2 = ns 2 = 10 x 28 = 14 σ 2 20 ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 242 NOTES Anna University Chennai Number of degrees of freedom = 10 – 1= 9 Table value of χ 2 for 9 df at 5% level = 16.919 H 0 is accepted since the calculated value of χ 2 < the table value of χ 2 . Hence the population variance can be regarded as 20 square kg.. Example 16: A random sample of size 20 from a normal population gives a sample mean of 42 and sample SD of 6. test the hypothesis that the population SD is 9. Clearly state the alternative hypothesis you allow for and the level of significance. Solution: H 0 : σ = 9 H 1 : σ | 9 s 2 = 6, σ 2 = 81 , n = 20 χ 2 = ns 2 = 20 x 36 =8.89 σ 2 81 Number of degrees of freedom = 20 – 1= 19 Table value of χ 2 for 19 df at 5% level = 30.144 H 0 is accepted since the calculated value of χ 2 < the table value of χ 2 . Therefore the population standard deviation can b regarded as 9. How you understood ? 1.Define Chi-square distribution? 2.State the important properties of χ 2 -distribution. 3.Give two uses of χ 2 distribution. 4.What is χ 2 - test of goodness of fit? 5.What is contingency table. TRY YOURSELF ! 1) The theory predicts that the proportion of beans in 4 given groups should be 9:3:3:1. In an examination with 1600 beans, the number in the 4 groups were 882, 313, 287 and 118. Does the experimental result support the theory? DIT 111 PROBABILITY AND QUEUEING THEORY 243 NOTES Anna University Chennai 2) 4 coins were tossed at a time and this operation is repeated 160 times. It is found that 4 heads occur 6 times, 3 heads occur 43 times, 2 heads occur 69 times and one head occurs 34 times. Discuss whether the coins be regarded as unbiased? 3) Five coins are tossed 256 times. The number of heads observed is given below. Examine if the coins are unbiased, by employing chi-square goodness of fit No: of heads 0 1 2 3 4 5 Frequency 5 35 75 84 45 12 4) 2 groups of 100 people each were taken for testing the use of vaccine. 15 persons contracted the disease out of the inoculated persons, while 25 contracted the disease in the other group. Test the efficiency of the vaccine . 5) An insurance company advertises in the press a special pension plan for self- employed persons. The advertisement includes a coupon which enables interested persons to complete and return to the company. The company then posts to the enquiries to the initial information about the pension plan. If there is no response from the enquiries to the initial information, a second information pack is sent to the enquiries. Enquiries are divided by the company into three categories: definitely takes on plan, shows interests in plan, not interested. The company analysed a sample of 200 respondents to the initial advertisement i.e., those who returned the coupon. The following data was obtained. Responds to I Responds to II Telephone mailing mailing call made Takes out plan 36 24 30 Shows interest 18 16 16 Not interested 6 20 34 Test whether there is any association between response and interest in the pension plan? 6) The heights of 10 randomly chosen college students in cm are 170, 165, 172, 168, 172, 164, 169, 167, 161, 163 Can we take the variance of heights of college students as 17 cm? DIT 111 PROBABILITY AND QUEUEING THEORY 244 NOTES Anna University Chennai REFERENCES: 1. T.Veerarajan, “Probability, statistics and Random Process “, Tata McGraw Hill, 2002. 2. P.Kandasamy, K. Thilagavathi and K. Gunavathi,”Probability, Random Variables and Random processors”, S. Chand, 2003. 3. S.C Gupta and V.K Kapoor,”Fundementals of Mathemetical Statistics”,Sultan Chand & Sons, 2002 DIT 111 PROBABILITY AND QUEUEING THEORY 245 NOTES Anna University Chennai UNIT 4 RANDOM PROCESSES - Introduction - Random process - Classification - Stationary process - Markov process - Markov chain - Poisson process 4. 1 INTRODUCTION In unit I you have studied about random variables. Random variable is defined as a function of the sample points in a simple space. It does not include the concept of time. But in the real world, we come across many time varying functions which are quite are random. By extending the concept of a random variable to include time it is possible to define a random process. In the case of random variables, only a real number is assigned to each sample point of the sample space of a random experiment. So we denoted it as X(s), a real function of s alone. But in case of random process, a function of time is assigned to each sample point, based on some rule.So it is denoted as X(s,t). In the simplest possible case a stochastic process amounts to a sequence of random variables known as a time series 4.2 LEARNING OBJECTIVES The students will acquire - Knowledge of random process concepts. - Skills in handling situations involving random variable when a function of time is assigned . DIT 111 PROBABILITY AND QUEUEING THEORY 246 NOTES Anna University Chennai - Knowledge in making scientific judgments in the time of uncertainty and variation. 4.3 RANDOM PROCESS 4. 3.1 Definition: A random process is a collection (or ensemble) of random variables {X(s,t)} that are functions of a real variable namely time t where s ε S (sample space) and t ε T (parameter set or index set) Another definition for random process: A random process or stochastic process is defined as a family of random variables {X(t)}defined at different instants of time. The set of possible values of any individual member of the random process is called state space. Any individual member itself is called a sample function or ensemble member or a realization of the process. i. If s and t are fixed {X(s,t)} is a number. ii. If t is fixed {X(s,t)}is a random variable. iii. If s is fixed, {X(s,t)}is a single time function. iv. If s and t are variables, {X(s,t)} is a collection of random variables that are time function. Hereafter if the parameter set T is discrete, the random process will be noted by {X(n)}or {X n }. If the parameter set T is continuous, the process will be denoted by {X(t)} 4.4 CLASSIFICATION Depending on the continuous or discrete nature of the state space and the parameter set T, a random process can be classified into four types: i) If both T and S are discrete, the random process is called a discrete random sequence. For example, if X n represents the outcome of the nth toss of a fair die, then {X n , n ¡ 1} is a discrete random sequence, since T = {1,2,3, . . .} and S = {1,2,3,4,5,6}. ii) If T is discrete and S is continuous, the random process is called a continuous random sequence. For example, If X n represents the temperature at the end of the nth hour of a day, then {X n , 1 , n , 24}is a continuous random sequence, since temperature can take any value in an interval and hence continuous. DIT 111 PROBABILITY AND QUEUEING THEORY 247 NOTES Anna University Chennai iii) If T is continuous and S is discrete, the random process is called discrete random process. For example, if X(t) represents the number of telephone calls received in the interval (0, t) then {X(t)} is a discrete random process, since S = {0,1,2,3,…..}. iv) If both T and S are continuous, the random process is called continuous random process. For example if x(t) represents the maximum temperature at a place in the interval (0, t), {X(t)} is a continuous random process. The word discrete or continuous is used to refer the nature of S and the word sequence or process to refer the nature of T. A random process is called a deterministic process if all the future values can be predicted from past observations. A random process is called a non-deterministic process if all the future values of any values of any sample function cannot be predicted from past observations. Probability Distribution and Density functions To each random variable we can define the probability distribution function F X (x 1: t 1 ) as F X (x 1: t 1 ) = P[X(t 1 ) , x 1 ] for any real number x 1 . This is called the first- order distribution function of random variable X(t 1 ). The first order probability density function f X (x 1 , t 1 ) is defined as the derivative of the first order probability density function. f X (x 1 , t 1 ) = d F X (x 1: t 1 ) dx 1 Joint Distribution For two random variables X(t 1 ) and X(t 2 ) defined at two time instants t 1 and t 2 from the random process X(t) we can define the second order joint probability distribution function as F X (x 1 ,x 2 :t 1 ,t 2 ) = P{X(t 1 ) , x 1 , X(t 2 ) , x 2 } and the second order joint probability density function as f X (x 1 ,x 2 : t 1, t 2 ) = c 2 F X (x 1 ,x 2 :t 1 ,t 2 ) cx 1 cx 2 We can extend this to n random variables. So n th order joint probability distribution is defined as F X (x 1 ,x 2, x n :t 1 ,t 2, ..... t n ) = P{X(t 1 ) , x 1 , X(t 2 ) , x 2 ,.... X(t n ) , x n } And the n th order joint probability density function as DIT 111 PROBABILITY AND QUEUEING THEORY 248 NOTES Anna University Chennai f X (x 1 ,x 2, x n :t 1 ,t 2, ..... t n ) = c n F X (x 1 ,x 2, x n :t 1 ,t 2, ..... t n ) cx 1 cx 2 ....... cx n Average values of Random process Mean of the process {X(t)} is the expected value of a typical member X(t) of t he process i.e µ = E(X(t)), Autocorrelation of the process {X(t)}, denoted by R xx (t 1 ,t 2 ) or R x (t 1 ,t 2 ) or R (t 1 ,t 2 ), is the expected value of the product of any two members X(t 1 ) and X(t 2 ) of the process. i.e., R (t 1 ,t 2 ) = E[X(t 1 ) × X(t 2 )] Autocovaraince of the process {X(t)}, denoted by C xx (t 1 , t 2 ) or C x (t 1 ,t 2 ) or C (t 1 ,t 2 ) is defined as C (t 1 ,t 2 ) = R (t 1 ,t 2 ) – µ(t 1 ) × µ(t 2 ) Correlation coefficient of the process {X(t)}, denoted by ρ xx (t 1 ,t 2 ) or ρ x (t 1 ,t 2 ) or ρ (t 1 ,t 2 ), is defined as ρ (t 1 ,t 2 ) = C (t 1 ,t 2 ) . \C(t 1 ,t 1 ) \C (t 2 ,t 2 ) Where C (t 1 ,t 2 ) is the variance of X(t 1 ) . 4.5 STATIONARITY 4.5.1 Stationary process: If certain probability distribution or averages do not depend on time (t), then the random process {X(t)} is called stationary. 4.5.1.1 Stationary to order one A random process is said to be stationary to order one if the first order density functions defined for all the random variables of the process are same. In other words first order density function of the process, should not change with a shift in time origin. f X (x 1 : t 1 ) = f X (x 1, t 1 + δ) should be true for any t 1 and a time-shift δ. As f X (x 1 : t 1 ) is independent of t 1 , obviously, the mean of each random variable is same. Hence the mean of the process is a constant. E[X(t)] = X = constant DIT 111 PROBABILITY AND QUEUEING THEORY 249 NOTES Anna University Chennai 4.5.1.2 Stationary to order two A random process is said to be stationary to order two if for all t 1 , t 2 and δ, its second order density function satisfy the condition f X (x 1 , x 2 : t 1 ,t 2 ) = f X (x 1 , x 2 : t 1 + δ, t 2 + δ) It is clear that, for a second- order stationary process , the second order joint probability function is a function of only time difference t 2 – t 1 and not on the absolute time. As it is possible to define first-order density functions of a second –process, is is clear that a second-order stationary process will also be a first order stationary process. For a second order stationary process the two moments E(X 1 2 ) and E(X 2 2 ) do not change with time and are constants whereas the second order moment E(X 1 X 2 ) = E(X(t 1 )X(t 2 )) = Rxx(t 1 ,t 2 ) which is also called as auto correlation function of the process is a function of time difference at which the random variable X 1 and X 2 are defined. E[X 1 X 2 ] = E[X(t 1 )X(t 2 )] = R XX (t 1 ,t 2 ) If t 2 – t 1 = τ, then R XX (t 1 ,t 2 ) = R XX (τ) 4.5.2 Strongly stationary process (SSS PROCESS) A random process is called a strongly stationary process or strict sense stationary process, if all its finite dimensional distributions are invariant under translation of time parameter. That is if the joint distribution of X(t 1 ), X(t 2 ),…… X(t n ) is the same as that of X(t 1 +h), X(t 2 +h),…… X(t n +h) for all t 1, t 2,. . . . t n and h > 0 and for all n=1, then the random process {X(t)}is called a SSS process. If the definition given above holds good for n = 1,2,…..k only and not for n>k, then the process is called kth order stationary. Two real valued random process {X(t)}and {Y(t)}are said to be jointly stationary in the strict sense, if the joint distribution of X(t) and Y(t) are invariant under translation of time. 4.5.3 Wide Sense Stationary Process (WSS PROCESS) A random process X(t) with finite first and second order moments is called weakly stationary process or covariance stationary process or wide- sense stationary process if its mean is a constant and auto correlation depends only on the time difference. i.e., if E{X(t)} = µ and E{ X(t) × X(t - τ) } = R(τ) DIT 111 PROBABILITY AND QUEUEING THEORY 250 NOTES Anna University Chennai Note: From the definitions given above, is clear that a SSS process with finite first order and second order moments is a WSS process, while a WSS process need not be a SSS process. Two real valued random process {X(t)}and {Y(t)}are said to be jointly stationary in the wide sense, if each process is individually a WSS process, R xy (t 1 ,t 2 ) is a function of (t 1 – t 2 ) only. Evolutionary process A random process that is not stationary in any sense is called an evolutionary process. Example 1:The process {X(t)}whose probability function is given by P{X(t) = n} = (at) n – 1 , n = 1,2,3,…. ( 1+ at) n + 1 = at . , n = 0 ( 1+ at) show that it is not stationary Solution: The probability distribution if X(t) is X(t) = n 0 1 2 3 ….. P n : at/(1 +at ) 1/(1 + at) 2 at/(1 + at) 3 (at) 2 /(1+ at) 3 · E(X(t)) = Σ np n n = 0 = 1/(1 + at) 2 + 2at/(1 + at) 3 + 3(at) 2 /(1 + at) 4 + . . . . . . = 1/(1 + at) 2 [1 + 2α + 3 α 2 +…..], where α = at/(1 + at) = 1/(1 + at) 2 [1 - α] -2 = [1/(1 + at) 2 ] (1 + at) 2 = 1 · · E(X 2 (t)) = Σ n 2 p n = Σ n 2 (at) n – 1 n = 0 n = 1 ( 1+ at) n + 1 · · = 1/(1 + at) 2 Σ n(n + 1)[at/1+at] n – 1 - Σ n[at/1+at] n - 1 n = 1 n = 1 · · = 1/(1 + at) 2 2 Σ n(n + 1)[at/1+at] n – 1 - Σ n[at/1+at] n - 1 n = 1 2 n = 1 ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 251 NOTES Anna University Chennai = 1/(1 + at) 2 2[1 - (at/1+at)] –3 - [ 1 – (at/1+at)] -2 · (since (1- x) -3 = Σ n(n + 1) x n-1 ) n = 1 2 = 2(1+ at) -3 - ( 1 + at) 2 = 2(1 + at) -1 ( 1 + at) 2 ( 1 + at) 2 E(X 2 (t)) = 1 + 2at V(X(t)) = E(X 2 (t)) – [E(X(t))] 2 = 1 + 2at – 1 = 2at since E(X(t)) and V(X(t)) are functions of t, {X(t)} is not stationary. Example 2:Examine whether the poisson process {X(t)}, given by the probability law P{X(t) = r} = e -λt (λt) r , r = 0, 1, …. is covariance stationary. r! Solution: The probability distribution of X(t) is a poisson distribution with parameter λt. Therefore E{X(t)} = λt | a constant. Therefore the poisson process is not covariance stationary. Example 3 : If {X(t)} ia wide sense stationary process with autocorrelation R(τ) = Ae -α|τ| , determine the second-order moment of random variable X(8) – X(5). Solution: Second-order moment of random variable X(8) – X(5) is given by E[{X(8) – X(5)} 2 ] = E{X 2 (8)} + E{X 2 (5)} – 2E{X(8)X(5)}. Given R(τ) = Ae -α|τ| i.e., R(t 1 , t 2 ) = Ae -α|t 1 – t 2 | Therefore E(X 2 (t)) = R(t, t) = A Therefore E{X 2 (8)} = E{X 2 (5)} = A Also E{X(8) X(5)} = R(8, 5) = Ae -3α Therefore E[{X(8) – X(5)} 2 ] = A + A - 2 Ae -3α E[{X(8) – X(5)} 2 ] = 2A(1 - e -3α ) DIT 111 PROBABILITY AND QUEUEING THEORY 252 NOTES Anna University Chennai Example 4:If X(t) = Y cosωt + Z sinωt, where Y and Z are two independent normal random variables with mean zero and same SDs and ω is a constant, prove that { X(t)} is a SSS process of order 2. Solution: Given E(X) = E(Y) = 0 and σ Y = σ Z = σ (say) Therefore Var(Y) = Var(Z) = σ 2 Since X(t) is a linear combination of Y and Z, that are independent, X(t) follows a normal distribution with E[X(t)] = cosωtE(Y) + sinωt E(Z) = 0 Var[X(t)] = cos 2 ωt × E(Y 2 ) + sin 2 ωt ×E(Z 2 ) Since X(t 1 ) and X(t 2 ) are each N(0, σ), X(t 1 ) and X(t 2 ) are jointly normal with the joint pdf given by f(x 1, x 2 , t 1 ,t 2 ) = 1 exp[-(x 1 2 – 2rx 1 x 2 + x 2 2 )/2(1-r 2 )σ 2 ], -· < x 1 ,x 2 <· 2 πσ 2 v(1 – r 2 ) where r = correlation coefficient between X(t 1 ) and X(t 2 ) = 1 E[X(t 1 ) . X(t 2 )] σ 2 = 1 E(Ycosωt 1 + Z sinωt 1 ) (Ycosωt 2 + Z sinωt 2 ) σ = 1 [E(Y 2 )cosωt 1 cosωt 2 + E(Z 2 ) sinωt 1 sinωt 2 ) σ 2 (E(YZ) = E(Y)E(Z) = 0) = 1 [σ 2 cosωt 1 cosωt 2 + σ 2 sinωt 1 sinωt 2 ) σ 2 = cosω(t 1 – t 2 ) In a similar manner the joint pdf of X(t 1 + h) and X(t 2 + h) can be expressed with R = cosω((t 1 + h) – (t 2 + h) ) = cosω(t 1 – t 2 ) Since the joint pdf of { X(t 1 ), X(t 2 )} and { X(t 1 + h) , X(t 2 + h)}are the same , {X(t)} is a SSS of order 2. DIT 111 PROBABILITY AND QUEUEING THEORY 253 NOTES Anna University Chennai Example 5: Show that the process X(t) = A cosλt + B sinλt (where A & B are random variables) is wide- sense stationary, if i)E(A) = E(B) = 0 ii) E(A 2 ) = E(B 2 ) and iii) E(AB) = 0. Solution: E(X(t)) = cosλt × E(A) + sinλt × E(B) (1) If {X(t)}is to be a WSS process, E(X(t)) must be a constant.(i.e independent of t). In (1) if E(A) and E(B) are any constants other than zero, E{X(t)} will be a function of t. Therefore E(A) = E(B) = 0 R(t 1 , t 2 ) = E{X(t 1 ) × X(t 2 )} = E{(A cosλt 1 + Bsinλt 1 )( A cosλt 2 + Bsinλt 2 )} = E(A 2 ) cosλt 1 cosλt 2 + B 2 sinλt 1 sinλt 2 + E(AB)sinλ(t 1 + t 2 ) (2) If {X(t)} is to be a WSS process, R(t 1 , t 2 ) must be a function of t 1 -t 2. Therefore in (2) E(AB) = 0 and E(A 2 ) = E(B 2 ) = u Then R(t 1 , t 2 ) = u cos λ(t 1 – t 2 ). Example 6: Consider a random process X(t) = A cos(ω 0 t + φ)Where A and φ are independent random variables and φ is uniformly distributed in the interval –π to π . Find the First and second moment of the process. Solution: As the random variables A and φ are independent f Aφ (a,) = f A (a). f φ (φ) φ is uniformly distributed in the interval –π to π f φ (φ) = 1/2π, – π < φ < π f Aφ (a,) = f A (a) (1/2π) DIT 111 PROBABILITY AND QUEUEING THEORY 254 NOTES Anna University Chennai First moment · π E(X(t)) = I I Acos(ω 0 t +φ) f A (a) (1/2π)dadφ -· - π · π = I A f A (a).da (1/2π) I cos(ω 0 t +φ)dφ -· - π · π = I A f A (a).da (1/2π) . 0 = 0 [ I cos(ω 0 t +φ)dφ = 0] -· -π First moment = 0 Second moment is · π E(X 2 (t)) = I I A 2 cos 2 (ω 0 t +φ) f A (a) (1/2π)dadφ -· - π · π = IA 2 f A (a).da (1/2π) I cos 2 (ω 0 t +φ)dφ -· -π The first integral is the mean square value of the random variable A. π So, E(X 2 (t)) = A 2 (1/2π) I (1 + cos 2(ω 0 t +φ))d φ [cos 2 θ = 1 + cos2θ ] - π 2 2 = A 2 (1/2π) ½ . 2π = ½ A 2 Example 7:For a random process X(t) = Y sin ωt, Y is a uniform random variable in the interval -1 to 1. Check whether the process is wide sense stationary or not.. Solution: To prove wide- sense stationary we have to prove the its mean is a constant and auto correlation depends only on the time difference. As Y is uniformly distributed in the interval -1 to 1 f Y (y) = ½ , -1 < y < 1 Mean of the process is 1 E{X(t 1 )}= I y sin ωt 1 f Y (y)dy -1 DIT 111 PROBABILITY AND QUEUEING THEORY 255 NOTES Anna University Chennai 1 = I y sin ωt 1 ½ dy -1 1 = sin ωt 1 ½ Iy dy -1 = 0 The autocorrelation of the process is E{X(t 1 ) × X(t 2 )} = E(Y sin ωt 1 Y sin ωt 2 ) = sin ωt 1 sin ωt 2 E(Y 2 ) 1 = sin ωt 1 sin ωt 2 Iy 2 f Y (y)dy -1 1 = sin ωt 1 sin ωt 2 Iy 2 (1/2)dy -1 = sin ωt 1 sin ωt 2 (1/3) =(1/3)cos(t 1 – t 2 ) - cos(t 1 + t 2 ) 2 Though the first moment namely mean is constant, the autocorrelation function is not a function of time difference of the two random variables alone. So the process is not wide sense stationary. Try yourself ! 1. For the sine wave process X(t) = Y cosω 0 t, -8 < t < 8, ω 0 = constant, the amplitude Y is a random variable with uniform distribution in the interval 0 to 1. Check whether the process is stationary or not. 2. Show that the random process X(t) = A cos (ω 0 t + φ) is wide sense stationary if A and ω 0 are constants and φ is uniformly distributed random variable in (0, 2π) 3. Given a random variable with characteristic function φ(ω) and a random process X(t) = cos( λt + Y). Show that {X(t)}is stationary in the wide sense φ(1) = 0 and φ(2) = 0. 4.6 MARKOV PROCESS AND MARKOV CHAIN A random process or stochastic process X(t) is said to be a Markov process if given the value of X(t), the value X(v) for v > t does not depend on the values of X(u) for u < t. In other words, the future behavior of the process depends only on the present value and not on the past values. DIT 111 PROBABILITY AND QUEUEING THEORY 256 NOTES Anna University Chennai 4.6.1 Markov Process A random process X(t) is said to be Markovian if P[X(t n+1 ) , x n +1 /X(t n ) = x n , X(t n-1 ) = x n-1 , ..X(t 0 ) = x 0 ] = P[X(t n+1 ) , x n+1 / X(t n ) = x n ] Where t 0 , t 1 , t 2 ,. . . . . , t n , t n+1 . X 0 , X 1 , X 2 , . . . X n , X n +1 are called the states of the process. If the random process at time at time t n is in the state X n , the future state of the random process X n+1 at time t n +1 depends only on the present state X n and not on the past states X n-1 ,X n-2 ,. . . X 0 . Examples of Markov process. 1.The probability of raining today depends on previous weather conditions existed for the last two days and not on past weather conditions. 2.A first order linear difference equation is Markovian. 4.6.1.1 Classification of Markov process. A Markov process can be classified into four types based on the nature of the values taken by t and {X i }. i) A continuous random process satisfying Markov property is called a continuous parameter Markov process as t and {X i } are both continuous. ii) A continuous random sequence satisfying Markov property is called a discrete parameter Markov process as the parameter t is discrete but {X i } is continuous. iii) A discrete random process satisfying Markov property is called a continuous parameter Markov chain as t is continuous and {X i }is discrete. iv) A discrete random sequence satisfying Markov property is called a continuous parameter Markov chain as t is discrete and {X i } is also discrete. 4.6.2 Markov Chain If for all n, P[X n = a n / X n-1 = a n – 1 , X n – 2 = a n-2 , . . . X 0 = a 0 ] = P[X n = a n / X n-1 = a n-1 ], then the process {X n }, n= 0,1,2, … is called Markov chain. (a 1 ,a 2 ,..a n ,…) are called the states of the Markov chain. The joint probability of Markov chain is P[X n = a n , X n-1 = a n – 1 , X n – 2 = a n-2 , . . . X 0 = a 0 ] = P[X n = a n / X n-1 = a n-1 ] P[X n = a n / X n-1 = a n – 1 ]…… P[X 1 = a 1 / X 0 = a 0 ]P[X 0 = a 0 ] DIT 111 PROBABILITY AND QUEUEING THEORY 257 NOTES Anna University Chennai n = P[X 0 = a 0 ] Π P [ X m = a m / X m – 1 = a m -1 ] m = 1 The conditional probability P[X n = a j / X n-1 = a i ] is called the one-step transition probability from state a i to state a j at the nth step and is denoted by p ij (n-1, n). If the one-step transition probability does not depend on the step i.e., p ij (n-1, n) = p ij (m-1, m) the Markov chain is called a Homogeneous Markov chain or the chain is said to have stationary transition probabilities. When the Markov chain is homogeneous, the one-step transition probability is denoted p ij . The matrix P = {p ij } is called (one-step) transition probability matrix (t p m). Note: The tpm of a Markov chain is a stochastic matrix, since p ij ¡ 0 and p ij = 1 j for all i, i.e., the sum of all the elements of any row of the tpm is 1. This is obvious because the transition from state a i to any state, including a i is a certain event. The conditional probability that the process is in state a j at step n given that it was in a state a i at step 0, i.e, P[X n = a j / X 0 = a i ] is called the n-step transition probability and denoted by p ij (n) . p ij (1) = p ij Let us consider an example in which we explain how the tpm is formed fro a Markov chain. Assume that a man is at an integral point of the x-axis between the origin and the point x = 3. He takes a unit step either to the right with probability 0.7 or to the left with probability 0.3, unless he is at the origin when he takes a step to the right to each x = 1 or he is at the point x = 3, when he takes a step to the left to reach x = 2. The chain is called ‘Random walk with reflecting barriers’ The tpm is given below States of X n 0 1 2 3 0 0 1 0 0 States of X n-1 1 0.3 0 0.7 0 2 0 0 0 0.7 3 0 0 1 0 Note: p 23 = the element in the 2 nd row, 3 rd row column of this tpm = 0.7. This means that, if the process is at state 2 at step (n -1), the probability that it moves to state 3 at step n = 0.7, where n is any positive integer. DIT 111 PROBABILITY AND QUEUEING THEORY 258 NOTES Anna University Chennai If the probability that the process is in state a i is p i (i = 1,2,…k) at any arbitrary step, then the row vector p = (p 1 ,p 2 , ….p k ) is called the probability distribution of the process at that time. In particular P (0) = {p 1 (0) , p 2 (0) ,.. . . . p k (0) }is the initial probability distribution, where p 1 (0) + p 2 (0) +.. . . .+ p k (0) = 1. [Remark: The transition probability matrix together with the initial probability distribution completely specifies a Markov chain {X n }. In the example given above, let us assume the initial probability distribution of the chain is P (0) = (¼, ¼,¼,¼) i.e., P{X 0 = i} = ¼, i = 0,1,2,3 Then we have, for example given below P{X 1 = 2 / X 0 = 1} = 0.7: P{X 2 = 1/X 1 = 2} = 0.3, P{X 2 = 1, X 1 = 2 / X 0 = 1}= P{X 2 = 1 / X 1 = 2}x P{X 1 = 2/ X 0 = 1}= 0.3 x 0.7 = 0.21 P{X 2 = 1, X 1 = 2 , X 0 = 1}= P{ X 0 = 1}x P{X 2 = 1, X 1 = 2/X 0 = 1} = ¼ x 0.21 = 0.0525 P{X 3 = 3, X 2 = 1, X 1 = 2, X 0 = 1}= P{X 2 = 1, X 1 = 2, X 0 = 1}x P{X 3 = 3 / X 2 = 1, X 1 = 2, X 0 = 1}= 0.0525 x 0 = 0 4.5.2.1Chapman-Kolmogorov Theorem If P is the tpm of a homogeneous Markov chain, then n-step tpm P (n) is equal to P n . i.e.,[P ij (n) ] = [P ij ] n . If p = {p i } is the state probability distribution of the process at an arbitrary time, then that after one step is pP, where P is the tpm of the chain and that after n step is pP n . A stochastic matrix P is said to be a regular matrix, if all the entries of p m (for some positive integer m) are positive. A homogeneous Markov chain is said to be regular if its tpm is regular. If a homogeneous Markov chain is regular, then every sequence of state probability distributions approaches a unique fixed probability distribution called the stationary distribution or steady-state distribution of the Markov chain. i.e., lim [p (n) ] = π, where the state probability distribution at step n, n÷· P (n) = {p 1 (n) , p 2 (n) ,.. . . . p k (n) }and the stationary distribution π = (π 1 , π 2 ,. . π k ) are row vectors. DIT 111 PROBABILITY AND QUEUEING THEORY 259 NOTES Anna University Chennai Note: If P is the tpm of the regular chain, then πP = π (π is a row vector). 4.6.2.2 Classification of states of a Markov chain: 1.State j is said to be accessible from state i if P ij (n) > 0 for some n ¡ 0. 2.Two states i and j which are accessible to each other are said to communicate i) state i communicates with itself for all i ¡ 0. ii) If state i communicates with state j and state j communicates with state k, then state i communicates with state k. 3.Two states that communicate are in the same class. Two classes of state are either identical or disjoint. If P ij (n) > 0 for some n and for all i and j, then every state can be reached from every other state. When this condition is satisfied, the Markov chain is said to be irreducible. The tpm of an irreducible chain is an irreducible matrix. Otherwise the chain is said to be non-irreducible or irreducible. A state is said to be an absorbing state if no other state is accessible from it; that is, for an absorbing state i, p ii = 1. State i of a Markov chain is called a return state, if p ii (n) > 0 for some n > 1. The period di of a return state i is defined as the greatest common divisor of all m such that p ii(m) > 0, i.e.,d i = GCD{m: p ii (m) > 0}. State i said to be periodic with period d i if d i > 1 and aperiodic if d i = 1. State i is aperiodic if p ii | 0. The probability that the chain returns to state i, having started from state i, for the first time at the nth step(or after n transitions) is denoted by f ii (n) and called the first return time probability or the recurrence time probability. {n, f ii (n) }, n = 1,2,3, …. is the distribution of recurrence times of the state i. · If F ii = f ii (n) = 1, the return to state i is certain. n = 1 · µ ii = nf ii (n) is called the mean recurrence time of the state i. n = 1 A state i is said to be persistent or recurrent if the return to state i is certain, i.e., if F ii = 1. DIT 111 PROBABILITY AND QUEUEING THEORY 260 NOTES Anna University Chennai The State i is said to be transient if the return to state i is uncertain, i.e., if F ii < 1. The state i is said to be non-null persistent if its mean recurrence time µ ii is finite and null persistent if µ ii = ·. A non-null persistent and aperiodic state is called ergodic. Theorem: If a Markov chain is irreducible, all its states are of the same type. They are all transient, all null persistent or all non-null persistent. All its states are either aperiodic or periodic with same period. Theorem: If a Markov chain is finite irreducible, all its states are non-null persistent Example 1: A raining process is considered as a two state Markov chain. If it rains, it is considered to be in state 0 and if does not rain, the chain is in state 1. The transition probability of Markov chain is defined as P = 0.6 0.4 0.2 0.8 Find the probability that it will rain for three days from today assuming that it is raining today. Find also the unconditional probability that it will rain after three days. Assume the initial probabilities of state 0 and1 as 0.4 and 0.6 respectively. Solution: The one-step transition probability matrix is given as P = 0.6 0.4 0.2 0.8 P 2 = 0.44 0.56 0.28 0.72 P 3 = 0.376 0.624 0.312 0.688 Therefore the probability that it will rain on third day given that it will rain today is 0.376 The unconditional probability that it will rain after three days is P(X 3 = 0) = 0.4P 00 3 + 0.6P 10 3 = (0.4)(0.376) + (0.6)(0.312) = 0.3376. Example 2: A person owning a scooter has the option to switch over to scooter, bike or a car next time with the probability of (0.3, 0.5, 0.2). If the transition probability matrix is 0.4 0.3 0.3 0.2 0.5 0.3 0.25 0.25 0.5 What are the probabilities related to his fourth purchase? DIT 111 PROBABILITY AND QUEUEING THEORY 261 NOTES Anna University Chennai Solution: 0.4 0.3 0.3 P = 0.2 0.5 0.3 0.25 0.25 0.5 P 3 = 0.277 0.351 0.372 0.269 0.359 0.372 0.275 0.345 0.380 Probabilities of his fourth purchase = (0.3, 0.5, 0.2)P 3 = (0.2726, 0.3538, 0.3736) Example 3: The transition probability matrix of a Markov chain {X n }, n= 1,2,…having 3 states 1,2 and 3 is P = 0.1 0.5 0.4 0.6 0.2 0.2 0.3 0.4 0.3 and the initial distribution p(0) = (0.7, 0.2, 0.1) Find i)P{X 2 =3} and ii) P{X 3 = 2, X 2 = 3, X 1 = 3, X 0 = 2}. Solution: P = 0.1 0.5 0.4 0.6 0.2 0.2 0.3 0.4 0.3 P 2 = 0.43 0.31 0.26 0.24 0.42 0.34 0.36 0.35 0.29 3 i) P{X 2 = 3} = P{X 2 = 3/X 0 = i}×P{X 0 = i} i =1 = P{X 2 = 3/X 0 = 1}×P{X 0 = 1} + P{X 2 = 3/X 0 = 2}×P{X 0 = 2} + P{X 2 = 3/X 0 = 3}×P{X 0 = 3} = p 13 (2) × 0.7 + p 23 (2) × 0.2 + p 33 (2) × 0.1 = 0.26 × 0.7 + 0.34 × 0.2 + 0.29 × 0.1 = 0.279 ii) P{X 3 = 2, X 2 = 3, X 1 = 3, X 0 = 2} = P{X 3 = 2/ X 2 = 3, X 1 = 3, X 0 = 2} × P{ X 2 = 3, X 1 = 3, X 0 = 2}.(by Markov property) DIT 111 PROBABILITY AND QUEUEING THEORY 262 NOTES Anna University Chennai = P{X 3 = 2/ X 2 = 3}× P{ X 2 = 3/ X 1 = 3, X 0 = 2}× P{X 1 = 3, X 0 = 2} = P{X 3 = 2/ X 2 = 3}× P{ X 2 = 3/ X 1 = 3}× P{X 1 = 3/X 0 = 2}× P{X 0 = 2} = p 32 × p 33 × p 23 × 0.2 = 0.4 × 0.3 × 0.2 × 0.2 = 0.0048. Example 3: Using limiting behaviour of Homogeneous chain, find the steady state probabilities of the chain given by the transition matrix 0.1 0.6 0.3 Probability P = 0.5 0.1 0.4 0.1 0.2 0.7 Solution: We know that, if P is the tpm of the regular chain, then πP = π and π = 1 0.1 0.6 0.3 [π 1 π 2 π 3 ] = [π 1 π 2 π 3 ] 0.5 0.1 0.4 0.1 0.2 0.7 π 1 = 0.1 π 1 + 0.5 π 2 + 0.1 π 3 π 2 = 0.6 π 1 + 0.1 π 2 + 0.2 π 3 π 3 = 0.3 π 3 + 0.4 π 2 + 0.7 π 3 π 1 + π 2 + π 3 = 1 Solving these equations π 1 = 0.2021 π 2 = 0.2553 π 3 = 0.5426 The steady state probability is (0.2021, 0.2553, 0.5426 ) Example 4: Find the nature of the states of the Markov chain with tpm 0 1 2 0 0 1 0 P = 1 ½ 0 ½ 2 0 1 0 Solution: 0 1 0 Given P = ½ 0 ½ 0 1 0 DIT 111 PROBABILITY AND QUEUEING THEORY 263 NOTES Anna University Chennai P 2 = ½ 0 ½ 0 1 0 ½ 0 ½ P = P 3 : P 4 = P 2 . In general P 2n = P 2 We note that p 00 (2) > 0 ; p 01 (1) > 0 ; p 02 (2) > 0 p 10 (1) > 0 ; p 11 (2) > 0 ; p 12 (1) > 0 p 20 (2) > 0 ; p 21 (1) > 0 ; p 22 (2) > 0 Therefore the Markov chain is irreducible. Also p ii (2) = p ii (4) = p ii (6) = . . . . . . = p ii (2n) = . . . . .> 0, for all i, all the states a re periodic and the period = GCD{2,4,6 . . . .} = 2 That is the period is 2. Therefore the chain is finite and irreducible, all its states are non-null persistent. All the states are not ergodic. Example 5: A man tosses a fair coin until 3 heads occur in a row. Let X n denotes the longest string of heads ending at the nth trial. Show that the process is Markovian. Find the transition matrix and classify the states. Solution: The state space = {0, 1, 2, 3}, since the coin is tossed until 3 heads occur in a row. The transition probability matrix is X n 0 1 2 3 0 ½ ½ 0 0 P = X n-1 1 ½ 0 ½ 0 2 ½ 0 ½ 0 3 0 0 0 1 Solution: Given ½ ½ 0 0 P = ½ 0 ½ 0 ½ 0 ½ 0 0 0 0 1 DIT 111 PROBABILITY AND QUEUEING THEORY 264 NOTES Anna University Chennai ½ ¼ ¼ 0 P 2 = ½ ¼ ¼ 0 ½ ¼ ¼ 0 0 0 0 1 ½ ¼ ¼ 0 P 3 = ½ ¼ ¼ 0 ½ ¼ ¼ 0 0 0 0 1 It is clear that the chain is irreducible. State 3 is absorbing. P ii (2) = p ii (3) = p ii (4) = . . . . . > 0 for all i d i = GCD{2,3,4, . . . . } = 1 Therefore all the states are aperiodic. Example 6: Consider a Markov chain with state space {0, 1} and the tpm P = 1 0 ½ ½ i) Draw a transition diagram. ii) Show that state 0 is recurrent. iii) Show that state 1 is transient. iv) Is the state 1 periodic? If so what is the priod? v) Is the chain irreducible. vi) Is the chain ergodic? Explain. Solution: i) DIT 111 PROBABILITY AND QUEUEING THEORY 265 NOTES Anna University Chennai Given P = 1 0 ½ ½ P 2 = 1 0 ¾ ¼ P 3 = 1 0 7/8 1/80 P 4 = 1 0 15/16 1/16 ii) A state i is said to be recurrent if and only if starting from state i, the process eventually returns to state i with probability one. Hence by the definition of recurrent, state 0 is recurrent. i) A sate i is said to be transient if and only if there is a +ve probability that the process will not return to this state. ii) We have f 11 (1) = ½ > 0 ; f 11 (2) = ¼ > 0 GCD (1, 2, 3, . . .. ) = 1 Hence the state 1 is periodic with period 1, i.e., aperiodic. iii) The chain is not irreducible as p 01 (n) = 0, n = 1,2,3,. . . .. iv) Since the chain is not irreducible, all the states are not non-null persistent. Hence the chain is not ergodic. Example 7: A fair die is tossed repeatedly. If X n denotes the maximum of the numbers occurring in the first n tosses, find the transition probability matrix P of the Markov chain {X n }. Find alsoP 2 and P(X 2 = 6) Solution: The state space is given by {1, 2, 3, 4, 5, 6}. Let X n = maximum of the numbers obtained in the first n trials = 4 (say) Then X n+ 1 = 4 if (n + 1)th trial is 1, 2, 3 or 4. = 5 if (n + 1)th trial is 5. = 6 if (n + 1)th trial is 6. P(X n +1 = 4 / X n = 4 ) = 1/6 + 1/6 + 1/6 + 1/6 = 2/3 P(X n +1 = i / X n = 4 ) = 1/6 , i = 5,6. DIT 111 PROBABILITY AND QUEUEING THEORY 266 NOTES Anna University Chennai Therefore the transition probability matrix is given by 1/6 1/6 1/6 1/6 1/6 1/6 0 2/6 1/6 1/6 1/6 1/6 0 0 3/6 1/6 1/6 1/6 P = 0 0 0 4/6 1/6 1/6 0 0 0 0 5/6 1/6 0 0 0 0 0 1 1 3 5 7 9 11 0 4 5 7 9 11 0 0 9 7 9 11 P 2 = 1 . 0 0 0 16 9 11 36 0 0 0 0 25 11 0 0 0 0 0 36 Initial state probabilities are P(X i = 0) = P(0) = 1/6, i = 1, . . . 6 6 Now P(X 2 = 6) = P(X 2 = 6 / X 0 = i) P(X 0 = i) i = 1 6 = 1/6 p i6 2 i = 1 = 1/6 × 1/36(11 + 11 + 11 + 11 + 11 + 36) = 91/26 Example 8: There are two white marbles in urn A and 3 red marbles in urn B. At each step of he process, a marble is selected from each urn and the two marbles selected are interchanged. Let the state ai of the system be the number of red marbles in A after 3 steps? In the long run, what is the probability that there are 2 red marbles in urn A? DIT 111 PROBABILITY AND QUEUEING THEORY 267 NOTES Anna University Chennai Solution: State space of the chain {X n } = {0, 1, 2}, since the number of balls in the urn A is always 2. Let the tpm of the chain {X n } be 0 1 2 0 p 00 p 01 p 02 P = 1 p 10 p 11 p 12 2 p 20 p 21 p 22 p 00 = 0 (since the state cannot remain at 0 after interchange of marbles) p 02 = p 20 = 0 (Since the number of red marbles in urn cannot increase or decrease by 2 in one interchange) To start with, A contains 0 red marble. After an interchange, A will contain1 red marble (and 1 white marble) certainly. Therefore p 01 = 1. Let X n = 1, i.e., A contains 1 red marble (and 1 white marble) and B contains 1 white and 2 red marbles. Then X n+1 = 0, if A contains 0 red marble (and 2 white marbles) and B contains 3 red marbles. i.e., if 1 red marble is chosen from A and 1 white marble is chosen from B and interchanged. Therefore P[X n+1 = 0 /X n = 1] = p 10 = ½ × 1/3 = 1/6 p 12 = ½ × 2/3 = 1/3 we know that p 10 + p 11 + p 12 = 1 Therefore p 11 = 1-( p 10 + p 12 ) = ½ p 21 = 2/3 p 22 = 1-(p 20 + p 21 ) = 1/3 Therefore the tpm is 0 1 0 P = 1/6 ½ 1/3 0 2/3 1/3 Now p(0) = (1, 0, 0) as there is no red marble in A in the beginning p (1) = p (0) P = (0, 1, 0) DIT 111 PROBABILITY AND QUEUEING THEORY 268 NOTES Anna University Chennai p (2) = p (1) P = (1/6, ½, 1/3 ) p (3) = p (2) P = (1/12, 23/26, 5/18) Therefore P[there are 2 red marbles in A after 3 steps] = P[X 3 = 2] = 5/18. Let the stationary probability distribution of the chain be π = (π 0 , π 1 , π 2 ) We know that, if P is the tpm of the regular chain, then πP = π and ? π = 1 0 1 0 (π 0 , π 1 , π 2 ) 1/6 ½ 1/3 = (π 0 , π 1 , π 2 ) 0 2/3 1/3 1/6 π 1 = π 0 π 0 + ½ π 1 + 2/3 π 2 = π 1 1/3 π 1 + 1/3 π 2 = π 2 π 0 + π 1 + π 2 = 1 Solving these equations we get π 0 = 1/10, π 1 = 6/10, π 2 = 3/10 P[there are 2 red marbles in A in the long run] = 0.3 Example 9: Three boys A,B and C are throwing a ball to each other. A always throw the ball to B and B always throws the ball to C, but C is just as likely to throw the ball to B as to A. Show that the process is Markovian. Find the transition matrix and classify the states. Solution: The transition probability matrix of the process {X n } is given below X n A B C A 0 1 0 P = B 0 0 1 X n-1 C ½ ½ 0 States X n depend only on states of Xn-1, but not on earlier states. Therefore {X n } is markovian chain. DIT 111 PROBABILITY AND QUEUEING THEORY 269 NOTES Anna University Chennai 0 0 1 P 2 = ½ ½ 0 0 ½ ½ ½ ½ 0 P 3 = 0 ½ ½ ¼ ¼ ½ p 11 (3) > 0, p 12 (1) > 0, p 13 (2) > 0, p 21 (2) > 0, p 22 (2) > 0, p 23 (1) > 0, p 31 (1) > 0, p 32 (1) > 0, p 33 (2) > 0. Therefore the chain is irreducible. 0 ½ ½ P 4 = ¼ ¼ ½ ¼ ½ ¼ ¼ ¼ ½ P 5 = ¼ ½ ¼ 1/8 3/8 ½ ¼ ½ ¼ P 6 = ¼ 3/8 ½ 1/8 3/8 3/8 and so on. Now p ii (2) > 0, p ii (3) > 0 , p ii (5) > 0, p ii (6) > 0 etc for i = 2,3, and GCD(2, 3, 5, 6, . . . .) = 1 Therefore the states 2 and 3 are periodic. i.e., the states A and B are periodic with period 1 i.e., aperiodic. Now p 11 (3) > 0 , p 11 (5) > 0, p 11 (6) > 0 etc and GCD(3,5,6, . . . . ) = 1 Therefore the state 1 is periodic with period 1 i.e., aperiodic. Since the chain is finite and irreducible, all its states and non-null persistent. Moreover all the states are ergodic. Try yourself ! 1) The tpm of a Markov chain {Xn}, n = 1,2,3, . . . with 3 states 0, 1, & 2 is ¾ ¼ 0 P = ¼ ½ ¼ 0 ¾ ¼ DIT 111 PROBABILITY AND QUEUEING THEORY 270 NOTES Anna University Chennai with initial distributions p(0) = (1/3, 1/3, 1/3). Find i)P(X 2 = 2) ii)P(X 3 = 1, X 2 = 2, X 1 = 2, X 0 = 2) (Solution: i) 1/6 ii) 3/64) 2) The one-step transition probability matrix of a Markovian chain with sate {0, 1} is given as P = 0 1 1 0 a)draw a transition diagram. b) Is it irreducible Markov chain? 3) Suppose that the probability of a dry day (state 0) following a rainy day (state 1) is 1/3 and that the probability of a rainy day following a dry day is ½ . Given that May 1 is a dry day, find the probability that i) May 3 is also a dry day ii) May 5 is also a dry day. (Solution: i)5/12 ii) 173/432 ) 4) A housewife buys 3 kinds of cereals, A, B and C. She never buys the same cereal in successive weeks. If she buys cereal A, the next wee she buys cereal B. However if she buys B or C the next week she is 3 times as likely as to buy A as the other cereal. How often she buys each of the three cereals. (Solution : (-3/7, 16/35 , 4/35)) 4.7 POISSON PROCESS In this section we consider an important example of a discrete random process kown as the Poisson process. 4.7.1 Definition: Let X(t) represents the number of occurrences of a certain event in (0, t) then the discrete random process {X(t): t = 0} is a Poisson process provided the following postulates are satisfied i) P(1 occurrence in (t, t + Δt)) = λ Δt + 0(Δt) ii) P(0 occurrence in (t, t + Δt)) = 1 – λ Δt + 0(Δt) iii) P(2 or more in (t, t + Δt)) = 0(Δt) iv) X(t) is independent of the number of occurrence of the event in any interval prior or after (0, t). v) The probability that the event occurs a specified number of times in (t 0 , t 0 + t) depends only on t but not on t 0 . Let the number of events in the interval [0, t] is denoted by X(t). Then the stochastic process {X(t): t = 0} is a Poisson process, with mean λ. Note that the number of events in the interval [0, t] is a Poisson distribution with parameter λt. DIT 111 PROBABILITY AND QUEUEING THEORY 271 NOTES Anna University Chennai The probability distribution of X(t) is given by P(X(t) = n) = e-λt(λt)n , n = 0, 1, 2, . . . . n! 4.7.2 Second-order probability function of a homogeneous Poisson process. P[X(t 1 ) = n 1 , X(t 2 ) = n 2 ] = P[X(t 1 ) = n 1 ]P[ X(t 2 ) = n 2 / X(t 1 ) = n 1 ] t 2 > t 1 = P[X(t 1 ) = n 1 ] P[the event occurs (n 2 – n 1 ) times in the interval length (t 2 – t 1 )] = e -λt1 (λt 1 ) n 1 e -λ(t2 – t1) [λ(t 2 – t 1 )] (n 2 –n1) , if n 2 ¡ n 1 n 1 ! (n 2 - n 1 )! = e -λt2 λ n 2 t 1 n 1 (t 2 – t 1 ) (n 2 –n1) if n 2 ¡ n 1 n 1 ! (n 2 - n 1 )! 0 otherwise Proceeding similarly, we can get the third-order probability function as P[X(t 1 ) = n 1 , X(t 2 ) = n 2 , , X(t 3 ) = n 3 ] = e -λt3 λ n3 t 1 n 1 (t 2 – t 1 ) (n 2 –n1) (t 3 – t 2 ) (n3 –n2) if n 3 ¡ n 2 ¡ n 1 n 1 ! (n 2 – n 1 )!(n 3 – n 2 )! 0 otherwise 4.7.3 Mean and Variance of the Poisson process. The probability law of the Poisson process {X(t)} is the same as that of a Poisson distribution with parameter λt. E{X(t)} = Var{X(t)} = λt. 4.7.4 Autocorrelation of the Poisson processs E(X 2 (t)) = λt + λ 2 t 2 . R xx (t 1 , t 2 ) = E[X(t 1 ) X(t 2 )] = E[X(t 1 ) {X(t 2 ) - X(t 1 ) + X(t 1 )}] = E[X(t 1 ) {X(t 2 ) - X(t 1 )}] + E {X 2 (t 1 )} = E[X(t 1 )]E[X(t 2 ) - X(t 1 )] + E {X 2 (t 1 )} since {X(t)} is a process of independent increments. = λt 1 [λ(t 2 – t 1 )] + λt 1 + λ 2 t 1 2 , if t 2 ¡ t 1 = λ 2 t 1 t 2 + λt 1 , if t 2 ¡ t 1 R xx (t 1 , t 2 ) = λ 2 t 1 t 2 + λt 1 , if t 2 ¡ t 1 DIT 111 PROBABILITY AND QUEUEING THEORY 272 NOTES Anna University Chennai 4.7.5 Auto covariance of the Poissson process C xx (t 1 , t 2 ) = R xx (t 1 , t 2 ) - E[X(t 1 )] E[ X(t 2 )] = λ 2 t 1 t 2 + λt 1 - λ 2 t 1 t 2 = λt 1 , if t 2 ¡ t 1 4.7.6 Correlation coefficient r xx (t 1 , t 2 ) = C xx (t 1 , t 2 ) . = λt 1 . = \(t 1 /t 2 ), if t 2 ¡ t 1 Var[X(t 1 )]Var[X(t 2 )] \(λt 1 λt 2 ) 4.7.7 Properties of Poisson process 1. The Poisson process is not a stationary process. The probability distribution of X(t) is a Poisson distribution with parameter λt. Therefore E{X(t)} = λt | a constant. Therefore the Poisson process is not covariance stationary. 2. The Poisson process is a Markov process. Proof: Consider P[X(t 3 ) = n 3 / X(t 2 )= n 2 , X(t 1 )= n 1 ] = P[X(t 1 )= n 1 ,X(t 2 ) = n 2 , X(t 3 )= n 3 ] P[X(t 1 )= n 1 ,X(t 2 ) = n 2 ] = e -λ(t3 - t2) λ (n 3 - n 2) (t 3 – t 2 ) (n 3 –n 2) (n 3 – n 2 )! (Refer section 4.4.2) = P[X(t 3 ) = n 3 / X(t 2 )= n 2 ] This means that the conditional probability distribution of distribution of X(t 3 ) given all the past values X(t 1 )= n 1 ,X(t 2 ) = n 2 depends only on the most recent value X(t 2 ) = n 2 . That is, the Poisson process possesses the Markov property. Hence the result. 3. The sum of two Poisson process is a Poisson process. Let {X 1 (t): t ¡ 0}, {X 2 (t): t ¡ 0} be 2 Poisson process and let X(t) = X 1 (t) + X 2 (t). n P(X(t) = n) = P(X 1 (t) = r) P(X 2 (t) = n - r) r = 0 n = e -λ1t (λ 1 t) r e -λ2 t [λ(t 2 )] (n - r) , if n 2 ¡ n 1 r = 0 r! (n - r)! DIT 111 PROBABILITY AND QUEUEING THEORY 273 NOTES Anna University Chennai = e -(λ1 + λ 2)t nCr(λ 1 t) r (λ 2 t) n -r n! = e -(λ1 + λ 2)t [(λ 1 + λ 2 )t] n n! which is a Poisson process with parameter (λ 1 + λ 2 )t. 4. The difference of two Poisson process is not a Poisson process. Proof: Let X(t) = X 1 (t) - X 2 (t). E(X(t)) = E[X 1 (t) - X 2 (t)] = E[ X 1 (t)] – E[X 2 (t)] E(X 2 (t)) = E[X 1 (t) - X 2 (t)] 2 = E[ X 1 2 (t)] – E[X 2 2 (t)] – 2E[X 1 (t)X 2 (t)] = (λ 1 t + λ 1 2 t 2 ) + (λ 2 t + λ 2 2 t 2 ) – 2(λ 1 t) (λ 2 t) = t 2 (λ 1 2 + λ 2 2 - 2 λ 1 λ 2 ) + (λ 1 + λ 2 )t = t 2 (λ 1 - λ 2 ) 2 + (λ 1 + λ 2 )t ? t 2 (λ 1 - λ 2 ) 2 + (λ 1 - λ 2 )t where E(X 2 (t)) for a Poisson process X(t) with parameters λ is given by λt + λ 2 t 2 . Therefore X(t) is a Poisson process. 5. The interarrival time of a Poisson process, i.e., the interval between two successive occurrences of a Poisson process with parameter λ has an exponential distribution with mean 1/λ. Proof: Let E i and E i+ 1 two consecutive events. Let E i take place at the time instant t i and T be the interval between the occurrences of E i and E i+ 1 .T is a continuous random variable. P(T > t) = P{ E i+ 1 did not occur in (ti, ti + t)} = P{No event occurs in an interval of length t} = P[X(t) = 0] = e -λt Therefore the cdf of T is given by F(T) = P {T , t} = 1 - e - λt Therefore the pdf of T is given by f(t) = λ e - λt (t ¡ 0) which is an exponential distribution with mean 1/λ . DIT 111 PROBABILITY AND QUEUEING THEORY 274 NOTES Anna University Chennai 6. If the number of occurrences of an event E is an interval of length t is a Poisson process {X(t)} with parameter λ and each occurrence of E has a constant probability p of being recorded and the recordings are independent of each other, then the number N(t) of the recorded occurrences in t is also a Poisson process with parameter λp. Proof: · P[N(t) = n] = P{E occurs (n + r) times in t and n of them are recorded} r = 0 · = e - λt (λt) n + r (n + r)C n p n q r , q = 1 - p r = 0 (n + r)! · = e - λt (λt) n + r (n + r)! p n q r r = 0 (n + r)! n! r! · = e - λt (λpt) n (λqt) r n! r = 0 r! = e - λt (λpt) n e λqt n! = e -λpt (λpt) n n! Example 1: Derive the probability law for the Poisson process. Solution: Let λ be the number of occurrences of the event in unit time. Let P n (t) = P{X(t) = n} Therefore P n (t + Δt) = P{X(t + Δt) = n} = P{(n - 1) calls in (0, t) and 1 call in (t, t + Δt)} + P{n calls in (0, t) and no calls in (t, t + Δt)} = P n – 1(t) λ Δt + P n (t) (1 - λ Δt) P n (t + Δt) - P n (t) = λ{ P n – 1(t) - P n (t)} Δt Taking the limits as Δt  0 d P n (t) = λ{ P n – 1 (t) - P n (t)} (1) dt Let the solution of the equation (1) be P n (t) = (λt) n f(t) (2) n! Differentiating (2) with respect to t, P | n (t) = λ n {nt n-1 f(t) + t n f | (t)} (3) DIT 111 PROBABILITY AND QUEUEING THEORY 275 NOTES Anna University Chennai Using (2) and (3) in (1) λ n t n f | (t) = -λ (λt) n f(t) n! n! i.e., f | (t) = -λf(t) f(t) = ke -λt (4) From (2), f(0) = P 0 (0) = P[ X(0) = 0] = P{no event occurs in (0, 0)} = 1. (5) Using (5) in (4), we get k = 1 and hence f(t) = e -λt (6) Using (6) in (2), Pn(t) = P( X(t) = n) = e -λt (λt) n , n = 0, 1, 2, . . . . n! Thus probability distribution of X(t) is given by P(X(t) = n) = e -λt (λt) n , n = 0, 1, 2, . . . . n! Example 2: If {X 1 (t)} and {X 2 (t)} are two independent Poisson processes with parameters λ 1 and λ 2 respectively, show that P[X 1 (t) = k / {X 1 (t) + X 2 (t) = n }] = nC k p k q n-k Where p = λ 1 and q = λ 2 . λ 1 + λ 2 λ 1 + λ 2 Solution: Required conditional probability = P[{X 1 (t) = k} 1 { X 1 (t) + X 2 (t) = n }] P{X 1 (t) + X 2 (t) = n } = P[{X 1 (t) = k} 1 {X 2 (t) = n - k }] P{X 1 (t) + X 2 (t) = n } = e -λ1t (λ 1 t) k × e -λ2t (λ 2 t) n -k k! (n -k)! e -(λ1 + λ2)t { (λ 1 + λ 2 )t} n n! (by independence and additive property) = n! (λ 1 t) k (λ 2 t) n -k k! (n - k)! { (λ 1 + λ 2 )t} n DIT 111 PROBABILITY AND QUEUEING THEORY 276 NOTES Anna University Chennai = nC k λ 1 k λ 2 . n -k λ 1 + λ 2 λ 1 + λ 2 = nC k p k q n-k Where p = λ 1 and q = λ 2 . λ 1 + λ 2 λ 1 + λ 2 Example 3: If {X(t): t ε T} is a Poisson process then the auto correlation coefficient between X(t) and X(t + s) is t . ½ (t + s) Solution: If {X(t): t ε T} be a Poisson process with parameter λ. Then E(X(t)) = λt, Var(X(t)) = λt E(X 2 (t)) = Var(X(t)) + [E(X(t)] 2 = λt + (λt) 2 By definition auto correlation is ρ(t, t + s) = C(t, t + s) [Var(X(t)) Var(X(t + s))] ½ C(t, t + s) = E[X(t) X(t + s)] – E(X(t))E(X(t + s)) Consider E(X(t))E(X(t + s)) = E[(X(t))(X(t + s)) – X(t) + X(t)] = E(X(t)X(t)) + E(X(t) X(t + s) – X(t)) = E(X(t) 2 ) + E(X(t))E (X(t + s) – X(t)) = E(X 2 (t)) + E(X(t))E (X(t + s) – E(X(t) 2 ) = λt + (λt) 2 + λt.λs Substituting this in C(t, t + s) we have C(t, t + s) = λt + (λt) 2 + λt.λs - λt.λ(t +s) = λt ρ(t, t + s) = λt . [ λt.λ(t +s)] ½ = t . (t 2 + ts) ½ = t . ½ (t + s) Example 3: Suppose the customers are arriving at a ticket counter according to a Poisson process with a mean rate of 2 per minute. Then in an arrival of 5 minutes find the probability that the number of customers arriving is i) exactly 3 ii) greater than 3 iii) less than 3. DIT 111 PROBABILITY AND QUEUEING THEORY 277 NOTES Anna University Chennai Solution: Given λ = 2 Therefore the number of customers X(t) arriving in an interval of duration t minutes follows a Poisson distribution with mean 2. i)The probability the number of customers arriving is exactly 3 is P[X(5) = 3] = e -10 (10) 3 3! ii)The probability the number of customers arriving is greater than 3 is P[X(5) > 3] = 1 - P[X(5) , 3] 3 = 1 - e -10 (10) k k = 0 k! iii)The probability the number of customers arriving is less than 3 is 2 P[X(5) < 3] = e -10 (10) k k = 0 k! Example 4: A machine goes out of order, whenever a component fails. The failure of this part follows a Poisson process with a mean rate of 1 per week. Find the probability that 2 weeks have elapsed since last failure. If there are 5 spare parts of this component in an inventory and that the ext supply is not due in 10 weeks, find the probability that the machine will not be out of order in the next 10 weeks. Solution: Here the unit time is 1 week. Mean failure rate = mean number of failures in a week = λ = 1 P(no failures in he2 weeks since last failure) = P[X(2) = 0] = e -λt (λt) n = e -2 (2) 0 = e -2 = 0.135 n! 0! There are only 5 spare parts and the machine should not go out of order in the next weeks 5 P[X(10) , 5] = e -10 10 n = 0.068s n = 0 n! Example 5: If customers arrive at a customer in accordance with a mean rate of 2 per minute, find the probability that the arrivals is i) more than 1 minute ii) between 1 minute and 2 minute and iii) 4 minute or less. DIT 111 PROBABILITY AND QUEUEING THEORY 278 NOTES Anna University Chennai Solution: The interval T between 2 consecutive arrivals follows an exponential distribution with parameter λ = 2. · P(T > 1) = I 2 e -t dt = e -2 = 0.135 1 2 i) P(1 < T < 2) = I 2 e -2t dt = e -2 – e -4 = 0.177 1 4 iii) P(T , 4) = I 2 e -2t dt = 1 – e -8 = 0.999 0 Example 6: A radioactive source emits a particle at a rate of 5 per minute in accordance with Poisson process. Each particle emitted has a probability 0.6 of being recorded. Find the probability that 10 particles are recorded in 4-min period. Solution: We know that If the number of occurrences of an event E is an interval of length t is a Poisson process {X(t)} with parameter λ and each occurrence of E has a constant probability p of being recorded and the recordings are independent of each other, then the number N(t) of the recorded occurrences in t is also a Poisson process with parameter λp. Here λ = 5 and p = 0.6 P(N(t) = k) = e -3t (3t) k k! P(N(4) = 10) = e -12 (12) 10 = 0.014 10! Example 7: The number of accidents in a city follows a Poisson process with a mean of 2 per day and the number X i of people involved in the i th accident has the distribution (independent) P{X i = k} = 1/2 k (k = 1). Find the mean and variance of the number of people involved in accidents per week. Solution: The mean and variance of the distribution P{X i = k} = 1/2 k , k = 1, 2, 3, . . . . .· can be obtained as 2 and 2. Let the number of accidents on any day be assumed as n. The numbers of people involved in these accidents be X 1 , X 2 , X 3 , . . . .X n DIT 111 PROBABILITY AND QUEUEING THEORY 279 NOTES Anna University Chennai X 1 , X 2 , X 3 , . . . .X n are independent and identically distributed random variables with mean 2 and variance 2. Therefore, by central limit theorem (X 1 + X 2 + X 3 + . . . + X n ) follows a normal distribution with mean 2n and variance 2n, i.e,. the total number of people involved in all the accidents on a day with n accidents = 2n. If N denotes the number of people involved in accidents on any day, then P(N = 2n) = P[X(t)) = n] where X(t) is the number of accidents. = e -2t (2t) n · 2! E(N) = 2n e -2t (2t) n n = 0 n! = 2E{X(t)} = 4t Var{N} = E{N 2 } – [E(N)] 2 · = 4n 2 e -2t (2t) n - 16t 2 n = 0 n! = 4E{X 2 (t)} - 16t 2 = 4{Var(X(t)} + [E(X(t))] 2 – 16t 2 = 4[2t + 4t 2 ] - 16t 2 = 8t Therefore, mean and variance of the number of people involved in accidents per week are 28 and 56 respectively. Example 8: If {X(t)} is a Poisson process, prove that P{X(s) = r/ X(t) = n} = nCr (s/t) r (1 – s/t) n – r where s < t Solution: P{X(s) = r/ X(t) = n} = P{X(s) = r}1{X(t) = n} P{X(t) = n} = P{X(s) = r}1{X(t - s) = n - r} P{X(t) = n} = P{X(s) = r}P{X(t - s) = n - r} (by independence) P{X(t) = n} = e -λs (λs) r × e -λ(t – s) (λ(t – s)) n -r r! (n - r)! e -λt (λt) n n! DIT 111 PROBABILITY AND QUEUEING THEORY 280 NOTES Anna University Chennai = n! s r (t - s) n – r r!(n -r)! t n = nCr (s/t) r (1 – s/t) n – r Example 9: If the particles are emitted from a radio active source at the rate of 20 per hour. Find the probability that exactly 5 particles are emitted during 15 minute period. Solution: Let X(t) denote the number of particles emitted during t minutes. By Poisson process P(X(t) = n) = e -λt (λt) n , n = 0, 1, 2, . . . . n! here λ = 20/60 = 1/3 P(X(t) = n) = e -t/3 (t/3) n , n = 0, 1, 2, . . . . n! P[exactly 5 particles are emitted during a 15 minute period] = P[X(15) = 5] = e -5 (5) 5 5! = 0.1755 Example 10: The probability that a person is suffering from cancer is 0.001. Find the probability that out of 4000 persons a) exactly 4 suffer because of cancer, b) more than 3 persons will suffer from the disease. Solution: The probability that a person is suffering from cancer = p = 0.001 No. of persons = 4000 Mean = λ = np = 4000 × 0.001 = 4 By the definition of Poisson distribution P[X = x] = e - λ λ x , x = 0, 1, 2, . . . . x! a)P[X = 4] = e -4 4 4 = 0.19532 4! 3 b)P[X > 3] = 1 – P[X = 3] = 1 - e -4 4 x x = 0 x! = 1 – 0.4333 = 0.5666 DIT 111 PROBABILITY AND QUEUEING THEORY 281 NOTES Anna University Chennai Try yourself ! 1. A radio active source emits particles at a rate 6 per minute in accordance with Poisson process. Each particle emitted has a probability 1/3 of being recorded. Find the probability that atleast 5 particles are recorded in a 5 minute period. (Solution: 0.9707 ) 2. If patients arrive at a clinic according to Poisson process with mean rate of 2 per minute. Find the probability that during a 1-minute interval, no patient arrives. (Solution: 0.135) 3. Suppose that customers arrive at a bank according to a Poisson process with a mean rate of 3 per minute; find the probability that during a time interval of 2 minutes i)exactly 4 customers arrive and ii)more than 4 customers arrive. (Solution: 0.133, 0.715) 4. Assume that the number of messages input to a communication channel in an interval of duration t seconds is a Poisson process with mean rate λ = 0.3. Compute i)the probability that exactly three messages will arrive during a ten-second interval. ii) The probability that the number of message arrivals ina interval of duration five seconds is between three and seven. (Solution: 0.224, 0.191) How you understood ? Say True 1. For a first-order stationary process, the density function of different random variables are different. 2. A wide-sense stationary process is a second-ordinary process. 3. A transition probability matrix of a finite sate Markov chain is a square matrix. 4. A stationary transition matrix of a finite state Markov chain always implies a stationary random sequence. 5. The transition probability matrix of a finite state Markov chain takes only non- negative values. 6. Poisson process is a discrete random process. 7. Poisson process is stationary 8. The interarrival time of a Poisson is also Poisson.. 9. Poisson process is Markovian. 10. The number of rows in a transition probability matrix of a finite state Markov chain is equal to the number of states of the system. (Answers: 1.false, 2.false, 3.true, 4.false, 5.true, 6.true, 7.false, 8.false, 9.true, 10.true) DIT 111 PROBABILITY AND QUEUEING THEORY 282 NOTES Anna University Chennai Short answer questions 1. What is a random process? 2. Classify Random process. Give example for each. 3. What is a first ordinary stationary process? 4. What is a wide-sense stationary process? Give an example. 5. Define a strict sense stationary process. 6. Classify Markov process. 7. Define Markov chain. 8. Define absorbing state, recurrent state and transient state of a Markov chain. 9. Give the properties of Poisson process. 10. Why a Poisson process is non-stationary ? REFERENCES: 1. T.Veerarajan, “Probability, statistics and Random Process”, Tata McGraw Hill, 2002. 2. P.Kandasamy, K. Thilagavathi and K. Gunavathi, “Probability, Random Variables and Random processors”, S. Chand, 2003. DIT 111 PROBABILITY AND QUEUEING THEORY 283 NOTES Anna University Chennai UNIT 5 QUEUEING THEORY - Introduction - Single and multi-server Markovian Queues - Little’s Formula - Average measures 5.1 INTRODUCTION Queuing or waiting lines arises in many situations of our daily life. For example passengers waiting for ticket booking ii) Machines wait for repair iii) patients waiting for treatment etc are different forms of queue being formed. If the queue is very long and customers in that have no patience to wait for service, they will seek other outlet. In such situations the servicing center may incur in loss. Therefore to solve the above problem the study of queuing theory is very important. Queuing theory analysis involves the study of systems behavior overtime. 5.2 LEARNING OBJECTIVES The students - will be exposed to basic characteristic features of a queuing system and acquire skills in analyzing queuing models. - Will have a well founded knowledge of different queuing models which can describe real life phenomena. 5.3 BASIC CHARACTERISTIC OF QUEUEING PHENOMENA 1) Input or arrival (inter-arrival distribution) 2) Output or departure (service) distribution 3) Service channels 4) Service discipline 5) Maximum number of customers allowed in the system 6) Calling service or population. DIT 111 PROBABILITY AND QUEUEING THEORY 284 NOTES Anna University Chennai  Customer: The units which arrive at a service centre are called customers.  Queue (waiting line): number of customers to be serviced  Queuing system: Number of customers in waiting line and also customers being serviced.  Service channel: the system that renders service to customers. The service channel may be single (one unit ) or multi channel (more than one). There may be one or more service stations or facilities at the service centre. If the service station is free, the customer who arrives there will be served immediately. If not the customer has to wait in line for his turn. Types of queuing models. There are several types of queuing models. Some of them are 1.Single queue – single service point. 2.Multiple queues – multiple service points 3.Simple queue – multiple service points 4.Multiple queues – single service point. The most common case of queuing models is the single channel waiting line. DIT 111 PROBABILITY AND QUEUEING THEORY 285 NOTES Anna University Chennai  Arrival rate (λ) is the number of customers arriving per unit of time.  Service rate (µ) is the number of customers served per unit of time.  Service discipline: is the order of service rule based on which customers in queue are serviced. For example : i) First come first served (FCFS) or First in first out (FIFO) as per which the customers are served in the strict order of their arrival. ii) If the last arrival in the system is served first, we have Last come first served (LCFS) iii) If the service is given in random order, Service in random order (SIRO) discipline. iv) General service discipline (GD) 5.4 OPERATING CHARACTERISTICS OF QUEUEING SYSTEM The operating characteristic of a queueing system refer to the numerical values of the probability distributions of various decision variables like arrival rate, number of facilities, service time, line length, priority system etc., some common characteristics area given below:  Queue length: Probability distribution of queue length can be obtained with the help of the given probability distribution of the arrival and service process. A large queue indicates poor service facility or a need for more space. On the other hand small queue indicates excess of service facilities.  Waiting time in the queue: It refers to the time spent by the customers in the queue before the commencement of his service. Long waiting time may increase the customer’s dissatisfaction and potential loss of future revenues. DIT 111 PROBABILITY AND QUEUEING THEORY 286 NOTES Anna University Chennai  Waiting time in system This is the total time spent by a customer in the queue plus service time. Long waiting time may indicate need for a change in the priority rules.  State of the system A basic concept in the analysis of a queueing theory is that of a state of the system. It involves study of a system’s behavior overtime. It is classified as follows:  Transient state A queueing system is said to be in transient state when its operating characteristics are dependent on time. A queueing system is in transient system when the probability distributions of arrivals, waiting time and service time of the customers are dependent on time. This state occurs at the beginning of the operation of the system.  Steady state If the operating characteristics become independent of time, the queueing system is said to be in a steady state. Thus a queueing system acquires steady state when the probability distribution of arrivals, waiting time and servicing time of the customers are independent of time. This state occurs in the long run of the system. 5.5 KENDALL’S NOTATION FOR REPRESENTING QUEUEING MODELS A general queueing system is denoted by (a/b/c) : (d/e) where a = probability distribution of the inter-arrival time b = probability distribution of the service time c = number of servers in the system d = maximum number of customers allowed in the system e = queue discipline Certain descriptive notations used for the arrivals and service time distributions i.e., to replace notation (a and b) are as follows. M = inter-arrival times or service times having exponential distributions, where the letter M stands for ‘Markovian’ property of the exponential. D = inter-arrival times or service times that are constant or deterministic. G = Service time distributions of general type i.e., no assumption is made about the form of distribution. Notations and symbols: The following symbols and notations will be used in connection with the queueing systems: n = total number of customers in the system, both waiting and in service. DIT 111 PROBABILITY AND QUEUEING THEORY 287 NOTES Anna University Chennai λ = Average number of customers arriving per unit of time µ = Average number of customers being served per unit of time s = number of parallel service channels (servers) L s or E(N s ) = E(n) = Expected or average number of customers in the system, both waiting and in service. L q = E(N q ) = E(m) = average or expected number of customers waiting in the queue (excluding those who are receiving the service) W s = E(W s ) = average or expected waiting time of a customer in the system both waiting and in service (including the service time). W q = E(W q ) = average or expected waiting time of a customer in the queue (excluding service time) P n (t) = probability that there are n customers in the system at any time t (both waiting and in service ) assuming that the system has started its operation at time zero. ρ = Traffic intensity or utilization factor which represents the proportion of time the servers are busy = λ/µ P n = steady state probability of having n customers in the system. 5.6 DIFFERENCE EQUATION RELATED TO POISSON QUEUE SYSTEM Let P n (t) be the probability that there are n customers in the system at time t (n > 0). Let us first derive the differential equation satisfied by P n (t) and deduce the difference equation satisfied by P n (probability of n customers at any time) in the steady-state. Let λ n be the average arrival rate when there are n customers in the system (both waiting in the queue and being served) and let µ n be the average service rate when there are n customers in the system. The system being in steady state does not mean that the arrival rate and service rate are independent of the number of customers in the system. Now P n ( t + Δt) is the probability of the n customers at time t + Δt. The presence of n customers in the system at time t + Δt can happen in any one of the following four mutually exclusive ways: DIT 111 PROBABILITY AND QUEUEING THEORY 288 NOTES Anna University Chennai i) Presence on n customers at t and no arrival or departure during Δt time. ii) Presence of (n -1) customers at t and one arrival and no departure during Δt time. iii) Presence of (n + 1) customers at t and no arrival and one departure during Δt time. iv) Presence of n customers at t and one arrival and one departure during Δt time (since more than one arrival/departure during Δt time is ruled out) Therefore P n (t + Δt) = P n (t)( 1 – λ n Δt) ( 1 – µ n Δt) + P n-1 (t) λ n-1 Δt ( 1 – µ n Δt) + P n+1 (t)( 1 – λ n+1 Δt) µ n+1 Δt + P n (t). λ n Δt .µ n Δt [since P(arrival occurs during Δt = λ Δt etc)] i.e., P n (t + Δt) = P n (t) – (λ n + µ n ) P n (t )Δt + λ n-1 P n-1 (t) Δt + µ n+1 P n+1 (t), on omitting terms containing (Δt) 2 which is negligibly small. P n (t + Δt) - P n (t) = λ n-1 P n-1 (t) – (λ n + µ n ) P n (t )Δt + µ n+1 P n+1 (t) (1) Δt Taking the limits on both sides of (1) as Δt  0, we have P n ’(t) = λ n-1 P n-1 (t) – (λ n + µ n ) P n (t )Δt + µ n+1 P n+1 (t) (2) Equation (2) does not hold good for n = 0, as P n- 1 (t) does not exist. Hence we derive the differential equation satisfied by P 0 (t) independently. Proceeding as before, P 0 (t + Δt) = P 0 (t) )( 1 – λ 0 Δt) + P 1 (t)( 1 – λ 1 Δt) µ 1 Δt, [by the possibilities (i) and (iii) given above and as no departure is possible when n = 0] P 0 (t + Δt) – P 0 (t) = -λ 0 P 0 (t ) + µ 1 P 1 (t) (3) Δt Taking limits on both sides of (3) as Δt  0, we have P 0 '(t) = -λ 0 P 0 (t ) + µ 1 P 1 (t) (4) Now in the steady state, P n (t)and P 0 (t) are independent of time and hence P n ’(t) and P 0 '(t) become zero. Hence the differential equations (2) and (4) reduce to the difference equations λ n-1 P n-1 – (λ n + µ n ) P n + µ n+1 P n+1 = 0 (5) and λ 0 P 0 + µ 1 P 1 = 0 (6) DIT 111 PROBABILITY AND QUEUEING THEORY 289 NOTES Anna University Chennai Value of P 0 and P n for Poisson Queue systems From equation (6) derived above, we have P 1 = λ 0 P 0 (7) µ 1 Putting n =1 in (5) and using (7), we have µ 2 P 2 = (λ 1 + µ 1 ) P 1 - λ 0 P 0 = (λ 1 + µ 1 ) λ 0 P 0 - λ 0 P 0 = λ 0 λ 1 P 0 µ 1 µ 1 Therefore P 2 = λ 0 λ 1 P 0 (8) µ 1 µ 2 Successively putting n = 2,3,. . . in (5) and proceeding similarly, we can get P 2 = λ 0 λ 1 λ 2 P 0 etc. µ 1 µ 2 µ 3 Finally P n = λ 0 λ 1 λ 2 . . . . λ n-1 P 0 etc. for n = 1,2, . . .. . . (9) µ 1 µ 2 µ 3 . . . . µ n since the number of customers in the system can be 0 or 1 or 2 or 3 etc, which events are mutually exclusive and exhaustive, we have · Σ P n n =1 · P 0 + Σ λ 0 λ 1 λ 2 . . . . λ n-1 P 0 = 1 n =1 µ 1 µ 2 µ 3 . . . . µ n Therefore P 0 = 1 (10)  1 + Σ λ 0 λ 1 λ 2 . . . . λ n-1 n =1 µ 1 µ 2 µ 3 . . . . µ n Equations (9) and (10) will be used to derive the important characteristics of the four queueing models. DIT 111 PROBABILITY AND QUEUEING THEORY 290 NOTES Anna University Chennai 5.7 CHARACTERISTICS OF INFINITE CAPACITY , SINGLE SERVER POISSON QUEUE MODEL I (M/M/1) : (/FIFO), when λ n = λ and µ n = µ (λ < µ) 1.Average number L s of customers in the system: Let N denote the number of customers in the queueing system (i.e., those in the queue and the one who is being served). N is a discrete random variable, which can take the value 0,1,2, . . . · Such that P(N = n) = P n = λ n P 0 , from equation (9) µ From equation (10), we have P 0 = 1 = 1 = 1 - λ · · µ 1 + Σ λ n Σ λ n n =1 µ n =0 µ P n = λ n 1 - λ µ µ · Now L s = E(N) = Σ n × P n n = 0 · = λ 1 - λ Σ n λ n -1 µ µ n = 1 µ = λ 1 - λ 1 - λ -2 , by binomial summation µ µ µ = λ = λ (1) . µ µ - λ 1 - λ µ 2. Average number L q of customers in the queue or average length of the queue: If N is the number of customers in the system, then the number of customers in the queue is (N -1). · Therefore L q = E( N - 1) = Σ (n – 1) × P n n = 1 · = 1 - λ Σ ( n -1) λ n µ n = 1 µ DIT 111 PROBABILITY AND QUEUEING THEORY 291 NOTES Anna University Chennai · = λ 2 1- λ Σ ( n -1) λ n - 2 µ µ n = 2 µ = λ 2 1 - λ 1 - λ -2 µ µ µ = λ 2 = λ 2 (2) . µ µ( µ - λ ) 1 - λ µ 3. Average number L w of customers in nonempty queues L w = E{(N - 1)/(N - 1) > 0}, since the queue is non-empty = E ( N -1) = λ × 1 . P(N – 1 > 0) (µ - λ ) · Σ P n n = 2 = λ 2 × 1 . µ( µ - λ ) · λ n 1 - λ Σ µ µ n = 2 = λ 2 × 1 . µ( µ - λ ) λ 2 1 - λ · λ n µ µ Σ µ n = 0 = µ × 1 . ( µ - λ ) 1 - λ 1 - λ -1 µ µ = µ (3) ( µ - λ ) 4. Probability that the number of customers in the system exceeds k · · P(N > k) = Σ P n = Σ λ n 1 - λ n = k +1 n = k +1 µ µ · = λ k +1 1 - λ Σ λ n – (k +1) µ µ n = k +1 µ DIT 111 PROBABILITY AND QUEUEING THEORY 292 NOTES Anna University Chennai · = λ k +1 1 - λ Σ λ n µ µ n = 0 µ = λ k +1 1 - λ 1 - λ -1 µ µ µ = λ k +1 (4) µ 5.Probability density function of the waiting time in the system Let W s be the continuous random variable that represents the waiting time of a customer n the system, viz, the time between arrival and completion of service. Let its pdf be f(w) and let f (w/n) be the density function of W s subject to the condition that there are n customers in the queueing system when the customer arrives, · Then f(w) = Σ f (w/n) P n (5) n = 0 Now f (w/n) = pdf of sum of (n +1) service times (one part-service time of the customer being served + n complete service times) = pdf of sum of (n +1) independent random variables, each of which is exponentially distributed with parameter µ = µ n +1 e -µw w n : w > 0 which is the pdf of Erlang distribution with n! parameter µ and n +1. · Therefore f(w) = Σ µ n +1 e -µw w n λ n 1 – λ by (5) n = 0 n! µ µ · = µ e -µw 1 – λ Σ 1 (λw) n µ n = 0 n! = µ 1 – λ e -µw e -λw , by exponential summation µ = (µ – λ) e -µw e –λw = (µ – λ) e –(µ –λ)w (6) which is the pdf of an exponential distribution with parameter (µ – ë). ¯ . ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 293 NOTES Anna University Chennai 6.Average waiting time of a customer in the system. W s follows an exponential distribution with parameter (µ – λ). E(W s ) = 1 (7) µ – λ (since the mean of an exponential distribution is the reciprocal of its parameter) 7.Probability that the waiting time of a customer in the system exceeds t · P( W s > t) = I f(w) dw t · = I (µ – λ) e –(µ –λ)w dw t · = (µ – λ ) e –(µ –λ) w –(µ – λ) t = e –(µ –λ) t (8) 8. Probability density function of the waiting time in the queue W q represents the time between arrival and reach of service point. Let pdf of W q be g(w) and let g(w/n) be the density function of W q subject to the condition that there are n customers in the system or there are (n -1) customers in the queue apart from one customer receiving service. Now g(w/n) = pdf of sum of n service times [one residual service time + (n -1) full service times] = µ n e -µw w n -1 : w >0 (n - 1)! · Therefore g(w) = Σ µ n e -µw w n -1 λ n 1 – λ n = 1 (n - 1)! µ µ · = λ 1 – λ e -µw Σ 1 (λw) n -1 µ n = 1 ( n - 1)! = λ (µ – λ) e -µw e –λw µ ¯ .¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 294 NOTES Anna University Chennai = λ (µ – λ) e –(µ –λ)w : w > 0 (9) µ and g(w) = 1 - λ , when w = 0 µ 9.Average waiting time of a customer in the queue · E(W q ) = λ (µ – λ) I w e –(µ – λ)w dw µ 0 · = λ I x e -x dx . µ 0 (µ – λ) · = λ [x (-e -x ) - e -x ] µ( µ - λ ) 0 = λ (10) µ( µ - λ ) 10. Average waiting time of a customer in the queue, if he has to wait E(W q / W q > 0) = E( W q ) P(W q > 0) = E( W q ) 1 - P(W q = 0) = E( W q ) . 1 - P(no customer in the queue) = E( W q ) 1 - P 0 = λ × µ (since P 0 = 1 - µ ) µ( µ - λ ) λ λ = 1 . ( µ - λ ) Little’s formula 1) E(N s ) = L s = λ . = λE(W s ) ( µ - λ ) DIT 111 PROBABILITY AND QUEUEING THEORY 295 NOTES Anna University Chennai 2) E(N q ) = L q = λ 2 . = λE(W q ) µ(µ - λ ) 3) E(W s ) = E(W q ) + 1 µ 4) E(N s ) = E(N q ) + λ µ If any one of the equations E(N s ), E(N q ), E(W s ) and E(W q ) is known, the other three can be found out using the relations given above. 5.8 CHARACTERISTICS OF INFINITE CAPACITY , MULTIPLE SERVER POISSON QUEUE MODEL II (M/M/s) : (/FIFO), when λ n = λ and µ n = µ (λ < sµ) 1.Values of P 0 and P n For the Poisson queue system P n is given by P n = λ 0 λ 1 λ 2 . . . . λ n-1 × P 0 n ¡ 1 (1) µ 1 µ 2 µ 3 . . . . µ n · -1 P 0 = 1 + Σ λ 0 λ 1 λ 2 . . . . λ n-1 (2) n =1 µ 1 µ 2 µ 3 . . . . µ n If there is a single server, µ n = µ for all n. But there are s servers working independently of each other, If there be less than s customers, i.e., if n < s, only n of the servers will be busy and the others idle and hence the mean service rate will be nµ. If n ¡ s, all the servers will be busy and hence the mean service rate = sµ. hence µ n = nµ, if 0 , n < s = sµ, if n ¡ s (3) Using (3) an (1) and (2), we have P n = λ n × P 0 , if 0 , n < s 1µ.2µ.3µ. . . .nµ = 1 λ n × P 0 , if 0 , n < s (4) n! µ and P n = λ n × P 0 {1µ.2µ.3µ. . . .(s -1)µ}{sµ.sµ.sµ. . . .(n - s +1) times } ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 296 NOTES Anna University Chennai P n = λ n Χ P 0 (s - 1)! µ s – 1 (sµ) n – s +1 P n = 1 λ n Χ P 0 , if n  s (5) s!s n-s µ · Now P 0 is given by Σ P n = 1 n = 0 s -1 · i.e., Σ 1 λ n + Σ 1 λ n × P 0 = 1 n = 0 n! µ n = 0 s!s n-s µ s -1 · i.e., Σ 1 λ n + s s Σ λ n × P 0 = 1 n = 0 n! µ s! n = s µs s -1 i.e., Σ 1 λ n + s s λ s 1 × P 0 = 1 n = 0 n! µ s! µs 1 - λ µs s -1 i.e., Σ 1 λ n + 1 λ s × P 0 = 1 n = 0 n! µ s! 1 - λ µs µs or P 0 = 1 . ()(6) s -1 Σ 1 λ n + 1 λ s n = 0 n! µ s! 1 - λ µs µs 2. Average number L q of customers in the queue or average length of the queue: · L q = E( N - s) = Σ (n – s) × P n x = s · = Σ x P x + s x = 0 · = Σ x × 1 λ s + x × P 0 x = 0 s!s x µ · = 1 λ s P 0 Σ x λ x s! µ x = 0 µs ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 297 NOTES Anna University Chennai = 1 λ s λ P 0 1 . s! µ µs 1 - λ 2 µs = λ s + 1 1 µ P 0 (7) s. s! 1 - λ 2 µs 3. Average number of customers in the system By 4 th Little’s formula E(N s ) = E(N q ) + λ µ = λ s + 1 1 µ P 0 + λ (8) s. s! 1 - λ 2 µ µs · This result can also be obtained by using the definition E (N s ) = Σ n P n n = 0 4. Average time a customer has to spend in the system By 1 st Little’s formula E(N s ) = λE(W s ) or E(W s ) = E(N s ) λ = λ s + 1 1 µ P 0 + λ s. s! 1 - λ 2 µ µs . λ = λ s 1 + 1 µ P 0 (9) µ µ s. s! 1 - λ 2 µs 5. Average time a customer has to spend in the queue By 2 nd Little’s formula ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 298 NOTES Anna University Chennai E(N q ) = λE(W q ) or E(W q ) = E(N q ) λ λ s + 1 = 1 µ P 0 s. s! 1 - λ 2 µs . λ λ s = 1 1 µ P 0 (10) µ s. s! 1 - λ 2 µs 6. Probability that an arrival has to wait Required probability = Probability that there are s or more customers in the system P(W s > 0) = P(N ¡ s) · · = Σ P n = Σ 1 λ n P 0 n = s n = s s! s n -s µ · = 1 . λ s P 0 Σ λ n - s s! µ n = s µs = λ s P 0 µ (11) s! 1 - λ µs 7. Probability that an arrival enters the service without waiting. Required probability = 1 – P(an arrival to wait) = 1 - λ s P 0 µ (12) s! 1 - λ µs 8. Mean waiting time in the queue for those who actually wait E(W q / W s > 0) = E( W q ) P(W s > 0) ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 299 NOTES Anna University Chennai Using (10) and (11) λ s s! 1 - λ = 1 1 µ P 0 × µs . µ s. s! 1 - λ 2 λ s P 0 µs µ = 1 (13) µs – λ 9. Probability that there will be someone waiting Required probability = P(N ¡ s + 1) · · = Σ P n = Σ P n – P( N = s) n = s + 1 n = s using (10) and (5) λ s λ s = µ P 0 – µ P 0 s! 1 - λ s! µs λ s λ s = µ P 0 µ . (14) s! 1 - λ µs 10. Average number of customers (in non-empty queues), who have to actually wait L w = E(N q / N q ¡ 1) = E(N q ) / P( N¡ s) λ s +1 P 0 s! 1 - λ = µ µs . s. s! 1 - λ 2 λ S P 0 µs µ = λ µs . (15) 1 - λ µs 11) P( W > t) = e -µt 1 + (λ/µ) s [ 1 – e - µt(s – 1 – (λ/µ)) ] P 0 s! ( 1 - (λ/µs)) ( s -1 - (λ/µ)) ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 300 NOTES Anna University Chennai 5.9 CHARACTERISTICS OF FINITE CAPACITY , SINGLE SERVER POISSON QUEUE MODEL III [(M/M/1) : (k/FIFO) ] model 1. Values of P 0 and P n For the Poisson queue system P n = P(N = n) in the steady state is given by the difference equations, λ n-1 P n-1 – (λ n + µ n ) P n + µ n+1 P n+1 = 0 : n > 0 and λ 0 P 0 + µ 1 P 1 = 0: n = 0 This model represents the situations in which the system can accommodate only a finite number k of arrivals. If a customer arrives and the queue is full, the customer leaves without joining the queue. Therefore for this model, µ n = µ, n = 1, 2,3, . . . . and λ n = λ, for n = 0,1,2, . . . (k -1) 0, for n = k, k + 1, . . . Using these values in the difference equations given above, we have µP 1 = λP 0 (1) µ P n+1 = (λ + µ ) P n - λ P n-1 , for 1 , n , k -1 (2) µP k = λP k – 1 , for n = k (since P k+1 has no meaning) (3) From (1), P 1 = λ P 0 µ from (2) µ P 2 = (λ + µ ) λ P 0 - λ P 0 µ P 2 = λ 2 P 0 and so on µ In general, P n = λ n P 0 , for 0 = n = k -1 µ From (3) 1 , n , k -1 P k = λ λ k - 1 P 0 = λ k P 0 µ µ µ k Now Σ P n = 1 n = 0 { ¯ . ¯ . ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 301 NOTES Anna University Chennai k i.e., P 0 Σ λ n = 1 n = 0 µ 1 - λ k + 1 i.e., P 0 µ = 1 1 - λ µ which is valid for λ > µ 1 - λ µ , if λ | µ (4) 1 - λ k +1 µ P 0 = 1 . k + 1 , if λ = µ (5) 1 - λ since lim µ = 1 . (()() λ  1 1 - λ k +1 k + 1 µ µ λ n 1 - λ µ µ , if λ | µ (6) 1 - λ k +1 µ P n = 1 . k + 1 , if λ = µ (7) 2. Average number of customers in the system k E(N) = Σ nP n = 1 - λ k n = 0 µ Σ n λ n 1- λ k +1 n = 0 µ µ = 1 - λ λ k µ µ Σ d (x n ), where x = λ 1 - λ k +1 n = 0 dx µ µ ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . { { { } DIT 111 PROBABILITY AND QUEUEING THEORY 302 NOTES Anna University Chennai = 1 - λ λ µ µ d (1 - x k +1 ) 1 - λ k +1 dx ( 1 - x) µ = (1 - x) x (1 - x){-(k + 1)x k } + (1 – x k +1 ) 1 – x k +1 (1 - x) 2 = x [1 – (k + 1)x k + k x k +1 ] (1 - x) (1 – x k +1 ) = x(1 – x k+ 1 ) – (k + 1)(1 - x) x k+ 1 (1 - x) (1 – x k +1 ) = x - (k + 1) x k +1 1 - x 1 – x k +1 (k + 1) λ k +1 = λ - µ if λ | µ (8) µ - λ 1 – λ k + 1 µ k and E(N) = Σ n = k if λ = µ (9) n = 0 k + 1 2 3. Average number of customers in the queue. k E(N q ) = E(N - 1) = Σ (n -1) P n n = 1 k k = Σ n P n - Σ P n n = 0 n = 1 = E(N) – (1 – P 0 ) (10) By 4 th Little’s formula E(N s ) = E(N q ) + λ µ or E(N q ) = E(N s ) - λ µ which is true when the average arrival rate is λ throughout. Now we see that, in step (10), 1 – P 0 | λ , because the average arrival rate is λ as long as there is a vacancy in the µ queue and it is zero when the system is full. ¯ . ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 303 NOTES Anna University Chennai Hence we define the overall effective arrival rate, denoted by λ’ or λ eff . by using step (10) and Little’s formula as λ’ = 1 – P 0 or λ' = µ(1 – P 0 ) = λ(1 - P n ) (11) µ Thus we have (12) E(N q ) = E(N s ) - λ' µ which is the modified Little’s formula for this model. 4. Average waiting times in the system and in the queue By modified Little’s formulas, E(W s ) = 1 E(N s ) (13) λ' and E(W q ) = 1 E(N q ) (14) λ' where λ' is the effective arrival rate. 5.10 CHARACTERISTICS OF FINITE CAPACITY , SINGLE SERVER POISSON QUEUE MODEL IV [(M/M/s) : (k/FIFO) ] model 1. Values of P 0 and P n P n = λ 0 λ 1 λ 2 . . . . λ n-1 P 0 or n ¡ 1 (1) µ 1 µ 2 µ 3 . . . . µ n · -1 P 0 = 1 + Σ λ 0 λ 1 λ 2 . . . . λ n-1 (2) n =1 µ 1 µ 2 µ 3 . . . . µ n For this [(M/M/s) : (k/FIFO) ] model, λ n = λ, for n = 0, 1,2, k -1 0, for n = k, k + 1, . . . . µ n = nµ, for n = 0,1,2, . …, s-1 sµ, for n = s, s + 1, . . . . using these values of λ n and µ n in (2) and noting that 1 < s < k, we get P 0 -1 = 1 + λ + λ 2 + . . . . . + λ + λ s . 1!µ 2!µ 2 (s -1)!µ s -1 (s -1)!µ s -1 µs { { { } { DIT 111 PROBABILITY AND QUEUEING THEORY 304 NOTES Anna University Chennai + λ s+1 +. . . . . .+ λ k . (s -1)!µ s -1 (µs) 2 (s -1)!µ s -1 (µs) k – s – 1 s - 1 = Σ 1 λ n + λ s 1 + λ + λ 2 + . . . . . + λ k - s n = 0 n! µ s! µ s µs µs µs s – 1 k P 0 -1 = Σ 1 λ n + 1 λ s Σ λ n – s (3) n = 0 n! µ s! µ n = s µs 1 λ n P 0 , for n , s n! µ (4) P n = 1 λ n P 0 for s < n , k s! s n – s µ 0, for n > k 2. Average queue length or average number of customers in the queue k E(N q ) = E(N - s) = Σ (n - s) P n n = s k = P 0 Σ (n - s) 1 λ n (using (4)) s! n = s s n – s µ λ s k - s = µ P 0 Σ x λ x s! x = 0 µs λ s k - s = µ P 0 ρ Σ x ρ x -1 where ρ = λ s! x = 0 µs k - s = λ s P 0 ρ Σ d (ρ x ) µ s! x = 0 dρ = λ s P 0 ρ d 1 – ρ k – s + 1 µ s! dρ 1 – ρ = λ s P 0 ρ - (1 - ρ) (k – s + 1) ρ k – s + (1 – ρ k – s+1 ) µ s! (1 – ρ) 2 = λ s P 0 ρ - (k – s ) (1 - ρ) ρ k – s - (1 - ρ) ρ k –s + (1 – ρ k – s+1 ) µ s! (1 – ρ) 2 = P 0 λ s ρ - (k – s ) (1 - ρ) ρ k – s + 1 – ρ k – s (1 – ρ + ρ) µ s! (1 – ρ) 2 } ] ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . { ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 305 NOTES Anna University Chennai = P 0 λ s ρ [ 1 - ρ k – s - (k - s) (1 – ρ) ρ k – s ) ] (5) µ s! (1 – ρ) 2 where ρ = λ µ 3.Average number of customers in the system k E(N) = Σ n P n n = 0 s - 1 k = Σ n P n + Σ nP n n = 0 n = s s - 1 k k = Σ n P n + Σ (n – s)P n + Σ sP n n = 0 n = s n = s s - 1 k s -1 = Σ n P n + E(N q ) + s Σ P n - Σ P n n = 0 n = 0 n = s s -1 k = E(N q ) + s - Σ (s - n) P n since Σ P n = 1 (6) n = 0 n = 0 s -1 Obviously s – Σ (s - n) P n | λ , so that step (6) represents Little’s formula. n = 0 µ In order to make (6) to assume the form of Little’s formula, we define the overall effective arrival rate λ’ or λ eff as follows: s -1 λ ' = s – Σ (s - n) P n . µ n = 0 s -1 λ' = µ s – Σ (s - n) P n (7) n = 0 with this definition of λ’, step (6) becomes E(N) = E(N q ) + λ’ (8) µ which is the modified Little’s formula for this model ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 306 NOTES Anna University Chennai 4. Average waiting time in the system and in the queue By modified Little’s formula E(W s ) = 1 E(N) (9) λ' and E(W q ) = 1 E(N q ) (10) λ' Example 1: In a railway marshalling yard, goods trains arrive at a rate of 30 trains per day. Assuming that the inter-arrival time follows an exponential distribution and the service time (the time taken to hump a train) distribution is also exponential with a average of 36 minutes. Calculate i) Expected queue size (line length) ii) Probability that the queue size exceeds 10. If the input of trains increases to an average of 33 per day, what will be the change in iI and ii) ? Solution: (Model I) Arrival rate λ = 30 trains per day = 30 = 1/48 trains per minute 60 × 24 Average of service time is given, that means 1/ µ is given 1/ µ = 36 minutes Therefore µ = 1/36 trains per minute. i) Expected queue size (line length) = L S = λ . = 3 trains.(omit the negative sign) λ - µ ii) Probability that the queue exceeds 10, P(N >10) = (λ/µ) k +1 = (36/48) 11 = 0.042 Now if the input increase to 33 trains per day, then we have λ = 33 = 11/480 trains per minute and µ = 1/36 trains per minute. 60 × 24 i) Expected queue size (line length) = L S = λ . = 4.71 [ 5 trains. λ - µ ii) Probability that the queue exceeds 10, P(N >10) = (λ/µ) k +1 = ((11 ×36) / 480) 11 = 0.121 DIT 111 PROBABILITY AND QUEUEING THEORY 307 NOTES Anna University Chennai Example 2: Arrivals at a telephone booth are considered to be Poisson with an average time of 12 minutes between one arrival and the next. The length of a phone call assumed to be distributed exponentially with mean 4 min. a) Find the average number of persons waiting in the system. b) What is the probability that a person arriving at the booth will have to wait in the queue? c) What is the probability that it will take him more than 10 minutes altogether to wait for the phone and complete his call? d) Estimate the fraction of the day when the phone will be in use? e) The telephone department will install a second booth, when convinced that an arrival has to wait on the average for at the least 3 minutes for phone . By how much flowof arrivals should increase in order to justify a second booth? f) What is the average length of the queue that forms time to time? Solution: (Model I) Mean inter arrival time = 1/λ = 12 minutes Therefore mean arrival rate, λ = 1/12 per minute Mean service time 1/µ = 4 minutes Therefore mean service rate µ = ¼ per minute a) E(N) = L S = λ . = 0.5 customer λ - µ b) P(W > 0) = 1 – P(W = 0) = 1 – P(no customer in the system) = 1 – P 0 = 1 – [1 – (λ/µ)] = λ/µ = 1/3 c) P(W > 10) = e -(µ - λ)×10 = e - (1/4 – 1/12 ) × 10 = 0.1889 d)P( the phone will be idle) = P(N = 0) = P 0 = 1 – (λ/µ) = 2/3. e)The second phone will be installed, if E(W q ) > 3 i.e., if = λ . > 3 µ(µ - λ) i.e., if = λ R . > 3 where λ R is the required arrival rate. µ(µ - λ R ) i.e., if = λ R . > 3 1/4(1/4 - λ R ) DIT 111 PROBABILITY AND QUEUEING THEORY 308 NOTES Anna University Chennai i.e., if λ R > ¾(1/4 - λ R ) i.e., if λ R > 3/16 - 3/4 λ R i.e., if λ R > 3/28 Hence the arrival rate should increase by 3/28 – 1/12 = 1/42 per minute, to justify a second phone. f) Average length of a non – empty queue L w = µ . = 1.5 persons µ - λ Example 3: A petrol pump station has 4 pumps. The service times follow the exponential distribution with a mean of 6 minutes and cars arrive for service in a Poisson process at the rate of 30 cars per hour. a) What is the probability that an arrival would have to wait in line? b) Find the average waiting time, average time spent in the system and the average number of cars in the system? c) For what percentage of time would a pump be idle on an average? Solution: (Model II) s = 4, λ = 30/hour 1/ µ = 6minutes = 6/60 hours = 1/10 hours µ = 10 / hour a) P(an arrival has to wait) = P(W > 0) = (λ/µ) s P 0 s!( 1 - (λ/µs)) = 3 4 × P 0 . = 13.5 × P 0 24 ×( 1 – ¾ ) S -1 -1 Now P 0 = Σ 1 (λ/µ) n + (λ/µ) s . n = 0 n! s!( 1 - (λ/µs)) -1 = ( 1 + 3 + ½ × 9 + 1/6 × 27) + 3 4 24 × (1 – 1/3) = 0.0377 { } DIT 111 PROBABILITY AND QUEUEING THEORY 309 NOTES Anna University Chennai Therefore P(W > 0) = 13.5 × P 0 = 13.5 × 0.0377 = 0.5090 b) E(W q ) = 1 1 × (λ/µ) s . × P 0 µ s × s! (1 - (λ/µs)) 2 = 1 × 3 4 × 0.0377 10 × 4 × 24 ( 1 – 3/4) 2 = 0.0509 hours E (W s ) = E(W q ) + 1/µ = 0.0509 hours ++ 1/10 hours = 0.1509 L s = λ E(W s ) = 30 × 0.1509 = 4.527 cars. c)The fraction of time when the pumps are busy = traffic intensity = (λ/µs) = 30/(10 ×4) = ¾ Therefore the fraction of time when the pumps are idle = ¼ Therefore, required percentage = 25% Example 4: There are three typists in an office. Each typist can type an average of 6 letters per hour. If the letters arrive for being typed at the rate of 15 letters per hour, a) What fraction of the time all the typists will be busy? b) What is the average number of letters waiting to be typed? c) What is the average time a letter has to spend for waiting and for being typed? d) What is the probability that a letter will take longer than 20 minutes waiting to be typed and being typed? Solution: (Model II) s = 3, λ = 15/hour µ = 6/ hour a)P(all the typists are busy) = P(there are 3 or more customers in the system) = P(N ¡ 3) DIT 111 PROBABILITY AND QUEUEING THEORY 310 NOTES Anna University Chennai = (λ/µ) s P 0 s!( 1 - (λ/µs)) = (15/6) 3 P 0 . 6 × (1 – (15/ 6×3)) S -1 -1 Now P 0 = Σ 1 (λ/µ) n + (λ/µ) s . n = 0 n! s!( 1 - (λ/µs)) -1 = ( 1 + 2.5 + ½ × ( 2.5) 2 ) + (2.5) 3 6 × (1 – 5/6) = 0.0449 Therefore P(N ¡ 3) = (15/6) 3 P 0 . 6 × (1 – (15/ 6×3)) = (15/6) 3 × 0.0449 . 6 × (1 – (15/ 6×3)) = 0.7016 b) E(W q ) = 1 × (λ/µ) s +1 . × P 0 s × s! (1 - (λ/µs)) 2 = 1 × (2.5) 4 × 0.0449 3 × 6 ( 1 – 2.5/3) 2 = 3.5078 c) E(W s ) = E(N s ) by Little’s formula λ = 1 E(N q ) + λ again by Little’s formula λ µ = 1 {3.5078 + 2.5} 15 = 0.4005 hours d) P( W > t) = e -µt 1 + (λ/µ) s [ 1 – e - µt(s – 1 – (λ/µ)) ] P 0 s! ( 1 - (λ/µs)) ( s -1 - (λ/µ)) { } ¯ . { } DIT 111 PROBABILITY AND QUEUEING THEORY 311 NOTES Anna University Chennai P( W > 1/3) = e -6 × 1/ 3 1 + (2.5) 3 [1 – e -(-2 × - 0. 5 ) ] × 0.0449 6 ( 1 - (2.5/3)) ( -0.5) = e -2 [ 1 + 0.7016 (1 - e) ] (-0.5) = 0.4616 Example 5: The local one-person barber shop can accommodate a maximum of 5 people at a time (4 waiting and 1 getting hair cut). Customers arrive according to a Poisson distribution with mean 5 per hour. The barber cuts hair at an average rate of 4 per hour. a) What percentage of time is barber idle? b) What fraction of the potentials customers are turned away? c) What is the expected umber of customers waiting for a hair-cut? d) How much time can a customer expect to spend in the barber shop? Solution: (Model III, λ | µ) λ = 5, µ = 4, k = 5 a) P(the barber is idle) = P(N =0) = P 0 = 1 - (λ/µ) 1 - (λ/µ) k +1 = 1 – 5/4 1 – (5/4) 6 = 0.0888 Percentage of time when the barber is idle = 8.88 [ 9 b)P(a customer is turned away ) = P(N > 5) = (λ/µ) n 1 - (λ/µ) 1 - (λ/µ) k +1 = (5/4) 5 1 - (5/4) 1- (5/4) 6 = 3125 = 0.2711 11529 Therefore, 0.2711potential customers are turned away. ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 312 NOTES Anna University Chennai c)E(N q ) = E(N) – (1 – P 0 ) = λ - (k + 1) (λ/µ) k+1 - (1 – P 0 ) µ - λ 1 - (λ/µ) k +1 = - 5 – 6 × (5/4) 6 - (1 – 0.0888) 1 – (5/4) 6 = 6 × (15625/4096) - 5.9112 = 2.2 customers. 11529/4096 d) E(W) = 1 E(N) λ' where λ' = µ( 1 – P 0 ) E(W) = 1 × E(N) = 3.1317 / 3.6448 = 0.8592 hours µ( 1 – P 0 ) Example 6: At a railway station, only one train is handled at a time. The railway yard is sufficient only fro two trains to wait, while the other is given signal to leave the station. Trains arrive at the station at an average rate of 6 per hour and the railway station can handle them on an average of 6 per hour. Assuming Poisson arrivals and exponential Poisson distribution, find the probabilities fro the numbers of trains in the system. Also find the average waiting time of a new train coming into the yard. If the handling rate is doubled, how will the above results get modified? Solution:(Model III ) i) λ = 6 per hour, µ = 6 per hour, k = 2 + 1 = 3 Here λ = µ, P 0 = 1 / k + 1 = ¼ P n = 1 / k + 1 = ¼ for n = 1,2,3 E(N) = k/2 = 1.5 E(W) = 1 E(N) λ' where λ' = µ( 1 – P 0 ) E(W) = 1 × E(N) µ( 1 – P 0 ) = 1 × 1.5 = 0.3333 hours = 20 minutes 6 × (3/4) ii)if the handling rate is doubled, ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 313 NOTES Anna University Chennai λ = 6 per hour, µ = 12 per hour, k = 3 λ | µ P 0 = 1 - (λ/µ) 1 - (λ/µ) k +1 = 1 – ½ 1 – (1/2) 4 = 0.5333 P n = (λ/µ) n 1 - (λ/µ) 1 - (λ/µ) k +1 = (8 / 15)(1/2) n , for n = 1,2,3. E(N) = λ - (k + 1) (λ/µ) k+1 µ - λ 1 - (λ/µ) k +1 = 1 – 4 × (1/2) 4 = 1 –(4/15) = 0.7333 trains 1 – (1/2) 4 E(W) = 1 E(N) λ' where λ' = µ( 1 – P 0 ) E(W) = 1 × E(N) µ( 1 – P 0 ) = 1 × (0.7333) = 0.13095 hours 12( 1 – (8/15)) Example 7: A 2-person barber shop has 5 chairs to accommodate waiting customers. Potential customers who arrive when all 5 chairs are full, leave without entering barber shop. Customers arrive at the average rate of 4 per hour and spend an average of 12 minutes in the barber’s chair. Compute P 0 , P 1 , P 7 , E(Nq) and E(W). Solution: (Model IV) λ = 4 per hour, µ = 5 per hour, s = 2, k = 2 + 5 = 7 DIT 111 PROBABILITY AND QUEUEING THEORY 314 NOTES Anna University Chennai S -1 k -1 a) P 0 = Σ 1 (λ/µ) n + (λ/µ) s Σ (λ/µs) n -s n = 0 n! s! n = s 1 7 -1 = Σ 1 (4/5) n + (1/2)(4/5) 2 Σ (2/5) n -2 n = 0 n! n = 2 -1 = 1 + 4/5 + 8/25{1 + 2/5 +(2/5) 2 + (2/5) 3 + (2/5) 4 + (2/5) 5 } = [ 9/5 + 8/25{ 1 – (0.4) 7 /1 – 0.4 }] -1 = 0.4287 b) P n = 1 λ n P 0 , for n , s n! µ Therefore P 1 = (4 /5) ×0.4287 = 0.3430 c) P n = 1 λ n P 0 , for s < n , k s!.s n-s µ P 7 = 1 × (4/5) 7 × 0.4287 2 × 2 7-2 = 0.0014 d) E(N q ) = P 0 (λ/µ) s ρ [1 – ρ k – s – (k - s)( 1 - ρ) ρ k -s ] ( where ρ = λ/µs) s! (1 - ρ) 2 = (0.4287).(0.8) 2 (0.4) [1 – (0.4) 5 – 5 × 0.6 ×(0.4) 5 ] 2 × (0.6) 2 = 0.15 customers S -1 e)E(N) = E(N q ) + s - Σ (s - n)P n n = 0 1 = 0.1462 + 2 - Σ (2 -n)P n n = 0 = 2.1462 – (2 × P 0 + 1 × P 1 ) = 2.1462 – (2 × 0.4287 + 1 × 0.3430) = 0.95 customers { } ¯ . ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 315 NOTES Anna University Chennai E(W) = 1 E(N) λ’ s-1 where λ’ = µ(s – Σ (s -n)P n ) n = 0 = 4[2 – (2 × 0.4287 + 1 × 0.3430)] = 3.1984 Therefore E(W) = 0.9458/3.1984 = 0.2957 hours Example 8 : A car servicing station has 2 bays where service can be offered simultaneously. Because of space limitation, only 4 cars are accepted for servicing. The arrival pattern is Poisson with 12 cars per day. The service time in both the bays is exponentially distributed with µ = 8 per bay. Find the average number of cars in the service station, the average number of cars waiting for service and the average time a car spends in the system. Solution: (Model IV) λ = 12 per day, µ = 8 per day, s = 2, k = 4 S -1 k -1 a) P 0 = Σ 1 (λ/µ) n + (λ/µ) s Σ (λ/µs) n -s n = 0 n! s! n = s -1 = 1 + 1.5/1 + ½ × (1.5) 2 {1 + (.75) + (.75) 2 } = 0.1960 E(N q ) = Average number of cars waiting for service = P 0 (λ/µ) s ρ [1 – ρ k – s – (k - s)( 1 - ρ) ρ k -s ] ( where ρ = λ/µs) s! (1 - ρ) 2 = (0.1960) × (1.5) 2 × (0.75) [1 – (0.75) 5 – 2 × 0.25 ×(0.75) 5 ] 2 × (0.25) 2 = 0.4134 car b) E(N) = Average number of cars in the system S -1 = E(N q ) + s - Σ (s - n)P n n = 0 1 = 0.4134 + 2 – Σ (2 - n)P n n = 0 { } ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 316 NOTES Anna University Chennai = 2.4134 – (2 P 0 + P 1 ) = 2.4134 – (2 × 0.1960 + 1.5 × 0.1960) = 1.73 cars c) E(W s ) = 1 E(N s ) λ’ s-1 where λ’ = µ(s – Σ (s -n)P n ) n = 0 = 8[2 – (2 P 0 + P 1 )] = 10.512 E(W s ) = 1.73/10.512 = 0.1646 day Example 9: In a given M/M/1 queueing system the average arrivals is 4 customers per minute : ρ = 0.7. What are 1) mean number of customers L s in the system 2) mean number of customers L q in the queue 3) probability that the server is idle 4) mean waiting time W s in the system. Solution: Given ρ = 0.7 (ρ = λ/µ) 1) L s = ρ . = 0.7 . = 2.333 1 – ρ 1 – 0.7 2)L q = ρ 2 = 1.6333 1 – ρ 3)P 0 = 1 – ρ = 1 – 0.7 = 0.3 4)W s = L s = 2.333/4 = 0.5833 λ Example 10:Customers arrive at a watch repair shop according to a Poisson process at a rate of one per every 10 minutes, and the service time is an exponential random variable with mean 8 minutes. 1) Find the average number of customers L s in the shop. 2) Find the average time a customer spends in the shop W s 3) Find the average number of customers in the queue L q 4) What is the probability that the service is idle. DIT 111 PROBABILITY AND QUEUEING THEORY 317 NOTES Anna University Chennai Solution: (Model I) Arrival rate λ = 1 per every 10 minutes Therefore λ = 1/10 Average of service time is given, that means 1/ µ is given 1/ µ = 8 minutes µ = 1/8 1) L S = λ . = 4 customer λ - µ 2) W s = 1 . = 40 minutes λ - µ 3)L q = L s - (λ / µ) = 4 – 8/10 = 3.2 ˜ 3 customers. 4)P 0 = 1 - (λ / µ) = 1 – (4/5) = 0.2 Example 11: A duplicating machine maintained for office use is operated by an office assistant. If the jobs arrive at a rate of 5 per hour and the time to complete each job varies according to an exponential distribution with mean 6 minutes, find the percentage of idle time of the machine in a day. Assume that jobs arrive according to a Poisson process. Solution: λ = 5 per hour 1/µ = 6 minutes = 6/60 hours µ = 10 per hour P(machine is idle) = P(N = 0) = 1 - (λ / µ) = ½ Therefore the percentage of idle time = 50% Example 12: What is the probability that an arrival to an infinite capacity 3 server Poisson process queue with λ / (cµ) = 2/3 and P 0 = 1/9 enters the service without waiting? Solution: P(Without waiting ) =1 – P(W > 0) = 1 - (λ/µ) s P 0 s!( 1 - (λ/µs)) = 2 3 × 1/9 . = 0.5556 6 ×( 1 – 2/3) DIT 111 PROBABILITY AND QUEUEING THEORY 318 NOTES Anna University Chennai Example 13: In a given M/M/1/ · /FCFS queue, ρ = 0.6, what is the probability that the queue contains 5 or more customers? Solution: P(X = 5) = (ρ) 5 = (0.6) 5 = 0.0467 Example 14: What is the effective arrival rate of M/M/1/4/FCFS queueing model when λ = 2 and µ = 5. Solution: Overall effective arrival rate = λ eff = λ(1 - P n ) = λ 1 - (λ/µ) n 1- (λ/µ) 1- (λ/µ) k +1 λ eff = λ(1 – P 4 ) = λ 1 - (λ/µ) 4 1- (λ/µ) 1- (λ/µ) 5 = 2 [1 - 2 4 ( 1-2 /1 -2 5 )] = 0.9677 Example 15: Consider an M/M/1 queueing system. If λ = 6 and µ = 8, find the probability of at least 10 customers in the system. Solution: · P(X = 10) = Σ P n = Σ(1 – 6/8)(6/8) n = (6/8) 10 = (3/4) 10 n = 10 Example 16: A bank has two tellers working on saving accounts. The first teller handles withdrawals only. The second teller handles deposits only. It has been found that the service time distributions for both deposits and withdrawals are exponential with mean service time of 3 minute per customer. Depositors are found to arrive in a Poisson fashion throughout the day with mean arrival rate of 16 per hour. Withdrawers also arrive in a Poisson fashion with mean arrival rate of 14 per hour. What would be the effect on the average waiting time for the customers if each teller could handle both withdrawals and deposits. What would be the effect, if this could only be accomplished by increasing the service time to 3.5 minutes? ¯ . ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 319 NOTES Anna University Chennai Solution: When there is a separate channel for the depositors, λ 1 = 16/hour, µ = 20/hour Therefore E(W q for depositors) = λ 1 . (Model I ) µ(µ - λ 1 ) = 16 . = 1/5 hours or 12 minutes 20 (20 - 16) When there is a separate Poisson channel for the withdrawers, λ 2 = 14/hour, µ = 20/ hour Therefore E( W q for withdrawers ) = λ 2 . (Model I ) µ(µ - λ 2 ) = 14 . = 7/60 hours or 7 minutes 20 (20 - 14) If both tellers do both service, (Model II) s = 2, µ = 20/hour, λ = λ 1 + λ 2 = 30 /hour E(W q ) = 1 1 × (λ/µ) s . × P 0 µ s × s! (1 - (λ/µs)) 2 = 1 × (1.5) 2 × P 0 20 × 2 × 2 ( 1 – 7.5) 2 = 0.45 × P 0 S -1 -1 Now P 0 = Σ 1 (λ/µ) n + (λ/µ) s . n = 0 n! s!( 1 - (λ/µs)) -1 = ( 1 + 1.5 + ( 2.5) 2 ) 2 × 0.25 = 1/ 7 Therefore E(W q ) = 0.45 × P 0 = 0.45 × 1/7 = 0.0643 hours Hence if both tellers do both types of service, the customers get benefited as their waiting time is considerably reduced. Now if both tellers do both types of service but with increased service time, s = 2, λ = 30, µ = 60/3.5 = 120/7 per hour { } DIT 111 PROBABILITY AND QUEUEING THEORY 320 NOTES Anna University Chennai E(W q of any customer) = 7 × 1 × (1.75) 2 × P 0 = 2.86 P 0 120 2 × 2 (1 – 7/8) 2 P 0 = 1 + 1.75 + ( 1.75 ) 2 -1 = 1/15 2 × 1/8 E(W q of any customer) = 2.86 P 0 = 2.86 × 1/15 = .1907 hours If this arrangement is adopted, withdrawers stand to lose as their waiting time is increased considerably and depositors get slightly benefited. How you understood ? 1. What are the characteristics of a queueing theory? 2. What do the letters in the symbolic representation (a/b/c):(d/e) of a queueing model represent? 3. What do you mean by transient state and steady state? 4. Write down the difference formulas for P 0 and P n in a Poisson queue system in the steady-state. 5. Write down the Little’s formulas that hold good for all the Poisson queue models. 6. Write down the formula for P n in terms of P 0 for the (M/M/s):(8/FIFO) queueing System. 7. How are Ns and Nq related in an (M/M/1):(k/FIFO) queueing system. 8. Define effective arrival rate with respect to (M/M/s):(8/FIFO) queueing System. TRY YOURSELF ! 1) Customers arrive in a one- man barber shop according to a Poisson process with a mean interval time of 12 minutes. Customers spend an average of 10 minutes in the barber’s chair. a) what is the expected number of customers in the barber shop an din the queue? b) Calculate the percentage of times an arrival can walk straight into the barber’s chair without having to wait? c) How much time can a customer expect to spend in the barber’s shop? d) Management will provide another chair and hire another barber, when a customer’s waiting time in the shop exceeds 1.25 hours. How much the average rate of arrivals increase to warrant a second barber? e) What is the average time customers spend in the queue? f) What is the probability that the waiting time in the system is greater than 30 minutes? g) Calculate the percentage of customers who have to wait prior to getting into the barber’s chair? h) What is the probability that more than 3 customers are in the system? (Solution: 4.17, 1/6, 1/300 per minute, 50minute, 0.6065, 83.33, 0.4823 ) ¯ . DIT 111 PROBABILITY AND QUEUEING THEORY 321 NOTES Anna University Chennai 2) At what average rate must a clerk in a super market work in order to ensure a probability of 0.90 that the customer will not wait longer than 12 minute? It is assumed that there is only one counter at which customers arrive in a Poisson fashion at an average rate of 15 per hour and that the length of the service by the clerk has an exponential distribution. (Solution: 24 customers per hour) 3) If the people arrive to purchase cinema tickets at the average rate of 6 perminute, it takes an average of 7.5 seconds to purchase a ticket. If a person arrives 2 minutes before the picture starts and it takes exactly 1.5 minute to reach the correct seat after purchasing the ticket, a) Can he expect to be seated from the starting of the picture? b) What is the probability that he will be seated from the starting of the picture? c) How early must he arrive in order to be 99% sure of being seated for the start of the picture? (Solution: 2 minutes, 0.63, 2.65 minutes) 4) Given an average arrival rate of 20 per hour, is it better for a customer to get service at a single channel with mean service rate of 22 customers per hour or at one of two channels in parallel with mean service rate of 11 customers per hour for each of the two channels. Assume both queues to be Poisson type. 5) A telephone company is planning to install telephone booths in a new airport. It has established the policy that a person should not have to wait more than 10% of the times he tries t use a phone. The demand for use is estimated be Poisson with an average of 0 per hour. The average phone call have an exponential distribution with a mean time of 5 minutes. How many phone booths should be installed? (Solution: 6 booths should be installed) 6) A supermarket has two girls attending the sales at the counters. If the service time for each customer is exponential with mean 4 minutes and if people arrive in Poisson fashion at the rate of 10 per hour, a) what is the probability that the customer has to wait for the service b) what is the expected percentage of idle time for each girl. c) if the customer has to wait in the queue, what is the expected length of his waiting time? ( Solution: 1/6, 67, 3 minutes) 7) Patients arrive at a clinic according to a Poisson distribution at a rate of 30 patients per hour. The waiting room does not accommodate more than 14 patients. Examination time per patient is exponential with mean rate of 20 per hour a) Find the effective arrival rate at the clinic. b) What is the probability that an arriving patient will not wait? DIT 111 PROBABILITY AND QUEUEING THEORY 322 NOTES Anna University Chennai c) What is the expected waiting time until a patient is discharged from the clinic? ( Solution: 19.98 per hour, 0.00076, 0.65 hours or 39 minutes) 8) A group of engineers has 2 terminals available to aid in their calculations. The average computing job requires 20 minutes of terminal time and each engineer requires some computation about once every half an hour. Assume that these are distributed according to an exponential distribution. IF there are 6 engineers in the group, find a) the expected number of engineers waiting to use one of the terminals and in the computing centre and b) the total time lost per day. ( Solution: 0.75, 16 × 0.0398 = 0.6368 hours) REFERENCES: 1. T.Veerarajan, “Probability, statistics and Random Process “, Tata McGraw Hill, 2002. 2. P.Kandasamy, K. Thilagavathi and K. Gunavathi,”Probability, Random Variables and Random processors”, S. Chand, 2003. 3. S.C Gupta and V.K Kapoor,”Fundementals of Mathemetical Statistics”, Sultan Chand & Sons, 2002 SUMMARY: At the end of this course you will be having fundenental knowledge of the basic probability concepts.You will be having well-founded knowledge of standard distributions which can describe real life phenomena. You would have acquiredskills in handling situations in involving more than one random and functions of random variable. You are exposed to basic characteristic features of a queueing system and would have acquired skills in analyzing queueing models.You are provided with necessary mathematical support and confidence to tackle real life problems.